A 1.20 kg copper rod resting on two horizontal rails 0.90 m apart carries a
current I = 55.0 A from one rail to the other. The coefficient of static friction
between the rod and rails is μs= 0.60.
(a) What is the smallest vertical magnetic field B that would cause the rod to
slide?
(b) Suppose a B field is directed at some angle to the vertical φ, with the current
along the rod directed into the page, as shown. Find an expression for B as a
function of φ for the case when the rod is just on the verge of beginning to slide.
(c) Find the value of φ which yields the smallest value of B that would cause
the rod to slide, together with the corresponding value of B.
Answers
(a) The smallest vertical magnetic field B that would cause the rod to
slide is 0.145 Tesla for given The coefficient of static friction
between the rod and rails is μs= 0.60
What is magnetic field ?A magnetic field is a vector field that describes the magnetic influence on moving charges, currents, and magnetic materials. A moving charge in a magnetic field is subjected to a force that is perpendicular to both its own velocity and the magnetic field.
(a) using formula
μs × m × g = I × L × B
μs= 0.60
M= 1.2 kg
I = current = 55.0 A
L = Length = 0.9 m
magnetic field (B) = 0.145 Tesla
(b) expression
force (f) = I × L × B × sinФ
(c) given B = 0.145 Tesla
μs × m × g= I × L × B × sinФ
Ф = 90°
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Related Questions
If 10 A of current flows through a 2 ohm resistor, what is the voltage of the battery?
20 V
0.2 V
OS V
12 V
Answers
The voltage of the battery would be 20 volts. Option I.
Voltage calculationAccording to Ohm's law, the voltage (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R). Mathematically,
V = I × R
In this case, the current (I) flowing through the resistor is given as 10 A and the resistance (R) of the resistor is given as 2 ohms. Substituting these values into the above formula, we get:
V = 10 A × 2 ohms = 20 volts
Therefore, the voltage of the battery is 20 volts.
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Two very large, nonconducting plastic sheets, each 10.0 cm
thick, carry uniform charge densities σ1,σ2,σ3
and σ4
on their surfaces, as shown in the following figure(Figure 1). These surface charge densities have the values σ1 = -7.30 μC/m2 , σ2=5.00μC/m2, σ3= 1.90 μC/m2 , and σ4=4.00μC/m2. Use Gauss's law to find the magnitude and direction of the electric field at the following points, far from the edges of these sheets.
A:What is the magnitude of the electric field at point A , 5.00 cm
from the left face of the left-hand sheet?(Express your answer with the appropriate units.)
B:What is the direction of the electric field at point A, 5.00 cm
from the left face of the left-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)
C:What is the magnitude of the electric field at point B, 1.25 cm
from the inner surface of the right-hand sheet?(Express your answer with the appropriate units.)
D:What is the direction of the electric field atpoint B, 1.25 cm
from the inner surface of the right-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)
E:What is the magnitude of the electric field at point C , in the middle of the right-hand sheet?(Express your answer with the appropriate units.)
F:What is the direction of the electric field at point C, in the middle of the right-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)
Answers
Answer:
Explanation:
To use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.
Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.
Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively. Since the electric field is perpendicular to the faces, the flux through them is zero. So, Q_in = (σ1 - σ2) * A, where A is the surface area of the curved part of the cylinder. Thus,
Φ_E = E * A = Q_in / ε0
E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C
The magnitude of the electric field at point A is 2.31 x 10^5 N/C.
B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.
The direction of the electric field at point A is RIGHT.
C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,
Φ_E = E * A = Q_in / ε0
E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C
The magnitude of the electric field at point B is 3.77 x 10^7 N/C.
D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.
The direction of the electric field at point B is LEFT.
E:
Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q
The net charge enclosed within this Gaussian surface is:
Q = σ1 × (2πrh)
where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:
Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C
Using Gauss's law, we can find the electric field at point C:
E × (2πrh) = Q/ε0
where ε0 is the permittivity of free space. Solving for E, we get:
E = Q / (2πε0rh)
Plugging in the values, we get:
E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C
Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.
To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.
The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:
qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)
where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.
Substituting the given values, we get:
qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC
Using Gauss's law, we have:
E * 2πrh = qenc/ε0
where ε0 is the permittivity of free space.
Solving for E, we get:
E = qenc / (2πrhε0) = 2.22 × 10^4 N/C
Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.
F:
The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.
The net flux of an electric field in a closed surface is directly proportionate to the charge contained, according to Gauss' equation.
State Gauss’s lawTo use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.
Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.
Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively.
Φ_E = E * A = Q_in / ε0
E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C
The magnitude of the electric field at point A is 2.31 x 10^5 N/C.
B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.
The direction of the electric field at point A is RIGHT.
C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,
Φ_E = E * A = Q_in / ε0
E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C
The magnitude of the electric field at point B is 3.77 x 10^7 N/C.
D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.
The direction of the electric field at point B is LEFT.
E:Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q
The net charge enclosed within this Gaussian surface is:
Q = σ1 × (2πrh)
where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:
Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C
Using Gauss's law, we can find the electric field at point C:
E × (2πrh) = Q/ε0
where ε0 is the permittivity of free space. Solving for E, we get:
E = Q / (2πε0rh)
Plugging in the values, we get:
E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C
Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.
To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.
The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:
qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)
where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.
Substituting the given values, we get:
qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC
Using Gauss's law, we have:
E * 2πrh = qenc/ε0
where ε0 is the permittivity of free space.
Solving for E, we get:
E = qenc / (2πrhε0) = 2.22 × 10^4 N/C
Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.
F:The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.
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(a) Find the frequency ratio between the two frequencies fi =256 Hz and f2 = 320 Hz. (b) Add the interval of a fifth to f2 to obtain fs, and find the frequency ratio fs/fi. (c) Find the frequency of f3.
Answers
(a) The frequency ratio between the two frequencies fi = 256 Hz and f2 = 320 Hz is:
[tex]\frac{f_2}{f_i} = \frac{320}{256} = \frac{5}{4} = 1.25[/tex]
So the frequency ratio is 1.25.
(b) Adding the interval of a fifth to f2 = 320 Hz gives:
fs = f2 * (3/2) = 320 * (3/2) = 480 Hz
The frequency ratio fs/fi is:
[tex]\frac{f_s}{f_i} = \frac{480}{256} = \frac{15}{8} = 1.875[/tex]
So the frequency ratio is 1.875.
(c) To find the frequency of f3, we need to add the interval of a fourth to f2:
f3 = f2 * (4/3) = 320 * (4/3) = 426.67 Hz
Therefore, the frequency of f3 is 426.67 Hz.
Two asteroids are suspended in space 50 meters apart. The masses of the asteroids are 2000000 kg and
3000000 kg.
Answers
Answer:
Explanation:
What is the gravitational force between them?
To calculate the gravitational force between two objects, we can use the formula:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (6.6743 x 10^-11 N * m^2 / kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them.
Plugging in the given values, we get:
F = (6.6743 x 10^-11 N * m^2 / kg^2) * (2000000 kg) * (3000000 kg) / (50 m)^2
F = 0.8046 N
Therefore, the gravitational force between the two asteroids is approximately 0.8046 N.
A Car accelerate Cuniformly from) 13 ms -1 to 31ms-1 while entering the motor way Covering the distance 220m
Answers
Answer:
3.84 m/s^2.
Explanation:
To solve this problem, we can use the following kinematic equation:
v^2 = u^2 + 2as
where:
v is the final velocity (31 m/s)
u is the initial velocity (13 m/s)
a is the acceleration (which is assumed to be constant)
s is the distance traveled (220 m)
We want to solve for the acceleration, so we can rearrange the equation as follows:
a = (v^2 - u^2) / 2s
Substituting the given values:
a = (31^2 - 13^2) / (2 x 220)
a = 3.84 m/s^2
Therefore, the acceleration of the car is 3.84 m/s^2.
please rate
The bigger the spring constant, the more__________the spring is.
Answers
The bigger the spring constant, the more stiff or rigid the spring is.
What does it signify when a spring's spring constant is higher?The exact amount of force needed to bend a spring depends on the spring constant. Although pounds/inch is a common measurement in North America, the standard international (SI) unit for spring constants is Newtons/meter. A stiffer spring has a greater spring constant, and vice versa.
What does it signify when the spring constant is higher?The exact amount of force needed to bend a spring depends on the spring constant. Although pounds/inch is a common measurement in North America, the standard international (SI) unit for spring constants is Newtons/meter. A stiffer spring has a greater spring constant, and vice versa.
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Select in the ticker-timer a frequency of 25 Hz or 50 Hz. Determine the period of the ticker-timer.
Answers
Answer:
The period of a ticker-timer is the time interval between two consecutive dots made by the ticker.
If the frequency of the ticker-timer is 25 Hz, then it makes 25 dots in one second. Therefore, the period of the ticker-timer can be calculated as:
Period = 1/frequency = 1/25 Hz = 0.04 seconds
If the frequency of the ticker-timer is 50 Hz, then it makes 50 dots in one second. Therefore, the period of the ticker-timer can be calculated as:
Period = 1/frequency = 1/50 Hz = 0.02 seconds
So, the period of the ticker-timer is 0.04 seconds for a frequency of 25 Hz and 0.02 seconds for a frequency of 50 Hz
Explanation:
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HELP ME!!!!If a researcher is designing an electromagnet for a life-saving medical application, which properties of the magnet will she need to take into account?
Select two answers!!
Wether or not magnetic field is constant.
Number of could of conducting wire.
Wether or not domains are present in iron core.
Metal composition of conducting wire.
Answers
Answer:
Number of coils of conducting wire and whether or not domains are present in iron core are the two properties of the electromagnet that the researcher will need to take into account.
Explanation:
The number of coils of conducting wire affects the strength of the magnetic field produced by the electromagnet. More coils will produce a stronger magnetic field, while fewer coils will produce a weaker magnetic field. The researcher will need to determine the appropriate number of coils to produce the desired strength of the magnetic field for the medical application.
The presence of domains in the iron core is also an important consideration. The iron core of the electromagnet helps to concentrate the magnetic field and increase its strength. The domains in the iron core align with the magnetic field produced by the current flowing through the wire, and this alignment reinforces the magnetic field. If the iron core does not have domains, the magnetic field produced by the electromagnet will be weaker. Therefore, the researcher will need to ensure that the iron core has domains to maximize the strength of the magnetic field for the medical application.
How have astronomers used models to explain galactic evolution through mergers and collisions? Use this model to explain how astronomers might test their understanding of the physical processes of the universe.
Answers
Answer:
Astronomers use computer models to simulate the process of galactic evolution through mergers and collisions. These models are based on our current understanding of the physical laws that govern the behavior of matter and energy in the universe. By running simulations of galactic mergers and collisions, astronomers can test their understanding of how these physical processes work in practice and how they contribute to the formation and evolution of galaxies.
One way that astronomers might test their understanding of the physical processes of the universe is by comparing the predictions of their models to observations of real galaxies. For example, if a model predicts that a particular type of galaxy should have a certain shape, size, or distribution of stars, astronomers can compare these predictions to observations of actual galaxies to see if they match up. If there is a discrepancy between the model's predictions and the observations, this can indicate that there are some physical processes that are not well understood or included in the model.
Another way that astronomers might test their understanding is by looking for patterns or trends in the properties of galaxies that are consistent with the predictions of their models. For example, if a model predicts that galaxies that have undergone a recent merger should have a particular distribution of gas and dust, astronomers can look for evidence of this pattern in observations of real galaxies. If they find that the predicted pattern is consistently observed in a large sample of galaxies, this can provide support for the model's predictions and the physical processes that it includes.
Overall, computer models of galactic evolution through mergers and collisions provide a powerful tool for astronomers to test their understanding of the physical processes of the universe. By comparing the predictions of their models to observations of real galaxies and looking for consistent patterns and trends, astronomers can refine their understanding of how galaxies form and evolve over time.
A current of O.S.A flows in a circuit with resistance 60 calculate the potential difference of the circuit
Answers
Therefore, the potential difference of the circuit is 30 volts.
What in electricity is a potential difference?The external effort required to move a charge from one position to another in an electric field is known as an electric potential difference, or voltage. A test charge that has an electric potential differential of +1 will experience a shift in potential energy.
To calculate the potential difference (V) of the circuit, we can use Ohm's Law, which states that V = IR, where I is the current flowing through the circuit and R is the resistance of the circuit.
In this case, the current (I) is given as 0.5 A and the resistance (R) is given as 60 Ω. Therefore, we can substitute these values into Ohm's Law to find the potential difference:
V = IR
V = 0.5 A × 60 Ω
V = 30 volts
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The voltage of a battery is V and the current is I. If the voltage is doubled to 2V, what is the new current?
O 1/4
O 21
O 1/2
041
Answers
Answer:The current in a lightbulb with a voltage of 35.0 V and a resistance of 175 ohm is 0.2 A.
Find the current in a lightbulb?
Given:
The voltage in a lightbulb is given by the equation V=IR
V is the voltage, I is current, and R is the resistance.
The voltage of the lightbulb is given as 35.0 V.
The resistance of the lightbulb is given as 175 Ohm.
As the equation is given,
V= IR
where I is current, R is resistance and V is the voltage.
Now, I = V/R
As the value of Voltage and resistance of the lightbulb is given, we will put in the above equation, we get;
I = 35.0/ 175 A
I = 0.2 A.
Hence, the current of the lightbulb is 0.2 A.
Therefore, Option C is the correct answer.
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Explanation:
A rock with a mass of 10.0 kg is balanced on top of a large boulder. Describe the forces acting on the rock, and use the concept of forces to explain why it stays on top of the boulder.
Answers
There are two forces acting on the rock: the force of gravity pulling it downward and the force of the boulder supporting it from underneath.
What is the force of gravity?The force of gravity is the gravitational attraction between the rock and the Earth. It pulls the rock downward with a force equal to its weight, which is given by the equation Fg = mg, where Fg is the force of gravity, m is the mass of the rock, and g is the acceleration due to gravity (approximately 9.81 m/s^2).
Why do boulder stays on top?The concept of forces explains why the rock stays on top of the boulder because the forces are balanced. The force of gravity pulling the rock downward is equal and opposite to the force of the boulder supporting it from underneath. As a result, the rock remains in equilibrium, or a state of balance, on top of the boulder. If either force were to change, the equilibrium would be disrupted, and the rock would either fall to the ground or be pushed off the boulder.
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What factors would create an ideal circuit?
Answers
An ideal circuit is a theoretical representation of an electrical circuit, where all components are perfect and all parameters such as resistance, capacitance, and inductance are zero.
The ideal circuit also has no energy losses, making it an ideal electrical system. To create an ideal circuit, the following factors must be considered:
1. Perfectly Conductive Wires: The wires and other conductors used in the circuit should be perfect conductors, which means the resistance should be zero. This will ensure that no energy is lost in the form of heat.
2. Zero Inductance: Inductance is a property of a circuit which causes a voltage drop when current flows through it. The ideal circuit should have no inductance so that the current can flow freely.
3. Zero Capacitance: Capacitance is a property in which electric charge builds up when current passes through it. To create an ideal circuit, the capacitance should be zero.
4. Zero Impedance: Impedance is the opposition to the flow of current in an electrical circuit. The ideal circuit should have zero impedance so that the current can flow freely.
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How did Newton discovered gravity?
Answers
Answer:
Isaac Newton, the English physicist, mathematician, and astronomer, discovered the concept of gravity in the late 17th century. The story of his discovery of gravity is one of the most famous in scientific history.
The most well-known anecdote is that Newton was sitting under an apple tree when an apple fell from the tree and hit him on the head. This event, however, is likely to be a myth created to make the story more memorable. Nonetheless, it is true that Newton began to wonder why objects fall to the ground instead of flying off into space.
Newton's curiosity led him to conduct experiments to understand the behavior of falling objects. He reasoned that the same force that caused an apple to fall to the ground was responsible for holding the moon in orbit around the Earth.
Newton's breakthrough came when he realized that the force that causes objects to fall to the ground is the same force that governs the motion of the planets in the solar system. He described this force as "gravity" and formulated his famous law of universal gravitation, which states that every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
Newton's discovery of gravity was a major scientific achievement that revolutionized our understanding of the physical world. It laid the foundation for the development of classical mechanics, and the law of gravitation has since been used to explain a wide range of phenomena in physics, from the motion of planets to the behavior of subatomic particles.
In summary, Newton discovered gravity through a process of curiosity, experimentation, and mathematical reasoning. Although the apple falling on his head is unlikely to be true, his discovery has had a profound impact on our understanding of the universe.
Answer:
Isaac Newton did not "discover" gravity, as it was already known that objects were attracted to each other. However, he did discover the law of universal gravitation, which states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of their separation distance.
Why do you think the pylon in Figure 24 is designed the way it is, and not in the way shown in Figure 25?
Answers
They are specifically made tο be ideal fοr cοnducting live electrical lines because οf their electrical insulatiοn and mechanical tοughness. A structure called an electric pylοn οf hοt-rοlled steel bevels οr gusset plates.
What kinds οf patterns are used tο create electrical pylοns?Other materials, such as cοncrete and wοοd, may alsο be utilised in additiοn tο steel. Transmissiοn tοwers can be divided intο fοur main categοries: suspensiοn, terminal, tensiοn, οr transpοsitiοn.
Whο was the electrical pylοn's designer?This Central Electricity Bοard held a cοmpetitiοn in 1927, and the winning entry was chοsen by the classical designer Sir Reginald Blοοmfield. He settled οn an A-frame structure with latticewοrk that was οffered by the American cοmpany Milliken Brοthers and is still in use tοday.
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Complete question:
A student uses 800 W microwave for three seconds how much energy does a student use
Answers
Answer:
The student use 2400 Joules
Explanation:
From the formula E = pt
p = 800W
t = 3 seconds
=> E = 800*3 = 2400J
A body moving at 50m/s decelerates uniformly at 2/ms? until it comes to rest. What distance does it cover from the time it starts to decelerate to the time it comes to rest.
Answers
Answer:
625
Explanation:
To solve this problem, we can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since the body comes to rest)
u = initial velocity (50 m/s)
a = acceleration (-2 m/s^2 since the body is decelerating)
s = distance
We want to find the distance (s) that the body covers from the time it starts to decelerate to the time it comes to rest. We can rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a)
Substituting the values we have:
s = (0^2 - 50^2) / (2 x (-2)) = 625 meters
Therefore, the body covers a distance of 625 meters from the time it starts to decelerate until it comes to rest.
When a ball is thrown into the air, its kinetic energy is lowest
A at its highest point.
B. at the moment it is released.
C. as it begins to fall back to the ground.
Answers
A porter can climb 10 staircase of 30cm each in 10 sec by carrying a 50kg bag. Calculate the power of the porter
Answers
Therefore, the power of the porter is 441,450 J/s, or approximately 441.5 watts.
What is work done?The work done by the porter in lifting the 50 kg bag up the stairs can be calculated as the product of the force applied and the distance moved.
The force applied is the weight of the bag, which is given by:
F = m * g
where m is the mass of the bag and g is the acceleration due to gravity, which is approximately 9.81 m/s². Substituting the given values, we get:
F = 50 kg * 9.81 m/s²
F = 490.5 N
The distance moved by the porter in lifting the bag up one staircase is 30 cm, and the porter climbs 10 staircases in 10 seconds, which gives a speed of:
v = (10 * 30 cm) / 10 s
v = 30 cm/s
The power of the porter is the rate at which work is done, which can be calculated as:
P = W / t
where W is the work done and t is the time taken. Substituting the values, we get:
P = F * d * v / t
P = 490.5 N * 10 * 30 cm * 30 cm/s / 10 s
P = 441,450 J/s
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Use the data in the table to determine the identities of the two gasses that you found could be components of water. Provide evidence to support your claim.
Answers
The two gases that could be components of water are indeed hydrogen and oxygen.
Evidence to support this claim:
1. The chemical formula for water is H2O, which means that it is composed of two hydrogen atoms and one oxygen atom.
2. The table of elements shows that hydrogen (H) and oxygen (O) are both elements that exist in nature.
3. The atomic mass of hydrogen (1.008) and oxygen (15.999) matches the molecular mass of water (18.015).
4. Water is produced when hydrogen gas (H2) is burned in the presence of oxygen gas (O2), according to the following equation: 2H2 + O2 → 2H2O.
Overall, the evidence supports the conclusion that hydrogen and oxygen are the two gases that could be components of water.
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An athlete whirls a 7.66 kg hammer tied to the end of a 1.4 m chain in a simple horizontal circle where you should ignore any vertical deviations. The hammer moves at the rate of 0.372 rev/s. What is the tension in the chain? Answer in units of N.
Answers
The hammer's centripetal acceleration is therefore 100.59 m/s².
Using an example, what is acceleration?An object has positive acceleration when it is going faster than it was previously. Positive acceleration was demonstrated by the moving car in the first scenario. Positive forward motion is being made by the car.
Hammer mass, m, is 6.55 kg. chain length, including the length of the arms, r = 1.3 m, Hammer's angular velocity is given by the formula: = 1.4 rev/s = 8.79646 rad/s (1 rev = 6.28 rad).
The formula a = V2/r, where V is the transverse velocity of the hammer, yields the centripetal acceleration.
V = r, hence
As a result, a = r²
A = 1.3 x 8.796462, or 100.59 m/s², is obtained by substituting the supplied numbers in the equation above.
The hammer's centripetal acceleration is therefore 100.59 m/s².
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A student uses 800 W microwave for 30 seconds how much energy does a student use
Answers
Answer:
The student used 24000 Joules of energy.
Explanation:
We can use the Energy Power equation to solve this example.
[tex]\sf E=Pt[/tex]
Where
[tex]\sf E[/tex] is the energy in Joules (J)
[tex]\sf P[/tex] is the power in Watts (W)
[tex]\sf t[/tex] is the time in seconds (s)
Numerical Evaluation
In this example we are given
[tex]\sf P=800\\t=30[/tex]
Substituting our given values into the equation yields
[tex]\sf E=800 \cdot 30[/tex]
[tex]\sf E=24000[/tex]
24000 Joules
[tex]\Large\bold{SOLUTION}[/tex]
To calculate the energy used by the student in this scenario, we can use the formula:
[tex]\sf{Energy\: (in\: Joules) = Power\: (in\: Watts) \times Time\: (in\: seconds)}[/tex]
Given that the student uses an 800 W microwave for 30 seconds, we can plug in these values to the formula:
[tex]\sf Energy = 800\: W \times 30\: s = 24,000\: J[/tex]
Therefore, the student uses 24,000 Joules of energy in this scenario.
[tex]\rule{200pt}{5pt}[/tex]
Owen hits a baseball with a velocity of 55 m/s. The ballpark fence is 120 m away.
Does the ball reach the fence if it leaves the bat traveling upward at an angle of 30°
to the horizontal?
Answers
Answer:
Explanation:
We can solve this problem using kinematic equations. We know that the initial velocity of the ball is 55 m/s at an angle of 30° to the horizontal. We can break this velocity into its horizontal and vertical components:
vx = v0 cos θ = 55 cos 30° = 47.6 m/s
vy = v0 sin θ = 55 sin 30° = 27.5 m/s
We can now use the vertical motion equation to find the time it takes for the ball to reach its maximum height:
Δy = vy t + 0.5 a t^2
At the maximum height, the vertical velocity of the ball is 0, so we have:
0 = vy + a t_max
Solving for t_max, we get:
t_max = -vy / a = -27.5 / (-9.8) = 2.81 s
The ball will take twice this time to reach the fence, since it needs to come back down to the ground:
t_total = 2 t_max = 5.62 s
The horizontal distance the ball travels during this time is:
Δx = vx t_total = 47.6 × 5.62 = 267.7 m
Since this distance is greater than the distance to the fence (120 m), the ball will reach the fence if it leaves the bat traveling upward at an angle of 30° to the horizontal.
In deep space, there is very little friction. Once they launch a probe into deep space, where there are no external forces acting on it, scientists shut the probe’s engines off because the scientists want the probe to
stop immediately.
speed up.
slow down.
move at constant velocity.
Answers
A crate of mass
m = 26 kg
rides on the bed of a truck attached by a cord to the back of the cab as in the figure below. The cord can withstand a maximum tension of 69 N before breaking. Neglecting friction between the crate and truck bed, find the maximum acceleration the truck can have before the cord breaks. (Enter the magnitude of the maximum acceleration in the forward direction.)
m/s2
Answers
Answer:
Explanation:
The maximum tension the cord can withstand is 69 N, so we know that the tension in the cord cannot exceed this value. The tension in the cord is related to the acceleration of the truck through Newton's second law:
ΣF = ma
where ΣF is the net force on the crate, m is the mass of the crate, and a is the acceleration of the truck.
In this case, the only force acting on the crate in the horizontal direction is the tension in the cord. Therefore, we can write:
ΣF = T = ma
where T is the tension in the cord.
We can solve this equation for the acceleration:
a = T/m
We know that the tension cannot exceed 69 N, so the maximum acceleration the truck can have before the cord breaks is:
a = 69 N / 26 kg
a ≈ 2.65 m/s^2
Therefore, the maximum acceleration the truck can have before the cord breaks is 2.65 m/s^2.
10. The energy states of an electron in a hydrogen atom is given by:
�
�
=
−
13.6
�
�
�
2
En=
n
2
−13.6 eV
Which of the following is not a possible energy of an emitted photon of the atom for an electron that is initially at
�
=
4
n=4?
0.66 eV
1.89 eV
2.55 eV
12.8 eV
Answers
Because it is less than the required minimum energy difference of 1.51 eV, the energy of 0.66 eV is not feasible. Hence, 0.66 eV is the correct answer.
When the hydrogen atom's energy in its ground state is 13.6 eV, what is the energy of the third excited state?The electron is first assumed to be in the ground state (n=1) in a hydrogen atom. Hence, the electron's energy in its ground state is 13.6 eV. This means that 12.75eV is needed to transfer electrons from the ground state to the third excited state.
The following equation provides the energy levels:
En = -13.6/n² eV
where n is the main quantum number.
An electron can move from the n=4 level to the n=3, n=2, or n=1 level after initialization. For each of these transitions, the relevant photon energies and energy differences are as follows:
n=4 to n=3: ΔE = En=3 - En=4 = (-13.6/3²) - (-13.6/4²) = 1.51 eV
n=4 to n=2: ΔE = En=2 - En=4 = (-13.6/2²) - (-13.6/4²) = 3.40 eV
n=4 to n=1: ΔE = En=1 - En=4 = (-13.6/1²) - (-13.6/4²) = 10.2 eV
As a result, the released photons could have energies of 1.51 eV, 3.40 eV, or 10.2 eV.
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Imagine that scientists placed a satellite at the Earth-Moon L1 Lagrangian
point, which is a point between Earth and the Moon where the gravitational
pulls from the two bodies are equal and opposite. What would happen if a
satellite at this position drifted slightly closer to Earth?
O A.
A. The gravitational pull from the Moon would correct the satellite
and bring it back to the Lagrangian point.
OB. The satellite would stop drifting and would remain fixed in this
position because of its tangential velocity.
OC. The satellite would continue to drift toward Earth as Earth's pull
became stronger than that of the Moon.
OD. The gravitational pull from the Sun would eventually pull the
satellite from this point and cause it to directly orbit the Sun.
Answers
Answer:
Explanation:
A. The gravitational pull from the Moon would correct the satellite and bring it back to the Lagrangian point.
At the Earth-Moon L1 Lagrangian point, the gravitational pulls from the Earth and the Moon are balanced, and the satellite is in a stable equilibrium. If the satellite drifted slightly closer to Earth, the gravitational pull from the Earth would become stronger, but the gravitational pull from the Moon would also increase due to its closer distance, and this would correct the satellite's motion and bring it back to the Lagrangian point.
Find the density of seawater at a depth where
I the pressure atm
at the
the
surface is 1050 kg/m³. Seawater has a bulk
modulus of 2.3 x 10° N/m². Bulk modulus is
defined to be
B =
Po AP
Ap
Answers
Answer:
To find the density of seawater at a certain depth, we need to use the following equation:
P = P0 + ρgh
where:
P0 = pressure at the surface (given as 1 atm = 101325 Pa)
ρ = density of seawater at the depth we're interested in
g = acceleration due to gravity (9.81 m/s^2)
h = depth below the surface
We also need to use the bulk modulus equation to find the change in pressure with depth:
B = (ρ/ρ0)(P-P0)/P
where:
ρ0 = density of seawater at the surface (given as 1050 kg/m^3)
P = pressure at the depth we're interested in
Combining these two equations, we get:
B = (ρ/ρ0)((P0 + ρgh) - P0)/P
B = ρgh/P
ρ = (BP)/(gh)
Substituting the given values, we get:
ρ = (2.3 x 10^9 N/m^2)(101325 Pa)/(9.81 m/s^2)(1050 kg/m^3)(1 atm)
ρ ≈ 1031.4 kg/m^3
Therefore, the density of seawater at a depth where the pressure is 1 atm and the density at the surface is 1050 kg/m^3 is approximately 1031.4 kg/m^3.
What are some examples of conservation of energy?
Answers
Answer:
power plant
collision
Battery
Burning wood
speaker
Beating drum
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A 4.0-kg mass is moving to the right at 3.0 m/s. An 8.0 kg mass is moving to the left at 2.0 m/s. If after collision the two
masses join together, what is their velocity after collision?
O-0.33 m/s
O-0.20 m/s
O +1.4 m/s
O +2.3 m/s
Answers
Answer:
- 0.33 m/s
Explanation:
An illustration is shown above,
In this case, since the two objects move in opposite directions before collision, then move together, the formula to be used is,
m1u1 - m2u2 = (m1 + m2)v
Where,
m1 = mass of the first object
u1 = initial velocity of the first object
v1 = final velocity of the first object
m2 = mass of the second object
u2 = initial velocity of the second object
v2 = final velocity of the second object
Therefore,
(4.0 • 3.0) - (8.0 • 2.0) = (4.0 + 8.0)v
12 - 16 = 12v
-4 = 12v
Divide both sides by 12,
-4 / 12 = 12v / 12
-1 / 3 = v
v = -0.33 m/s
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