Answer:
Explanation:
a. The X and Y components of the velocity can be found using trigonometry:
X = V * cos(θ) = 15 m/s * cos(38°) ≈ 11.63 m/s
Y = V * sin(θ) = 15 m/s * sin(38°) ≈ 9.14 m/s
b. The time the ball is in the air can be found using the Y component of the velocity and the acceleration due to gravity:
Y = V * sin(θ) * t - (1/2) * g * t^2
where g = 9.8 m/s^2 is the acceleration due to gravity
Solving for t, we get:
t = 2 * Y / g ≈ 1.87 s
c. The distance the ball travels can be found using the X component of the velocity and the time in the air:
distance = X * time = 11.63 m/s * 1.87 s ≈ 21.78 m
A power plant involves thermodynamic cycles to generate electrical power. In the first stage, water is pumped under saturated conditions from a pressure of 0.7 bar to 30 bar. Water then goes to the boiler at constant pressure and leaves the boiler at 500°C. In this condition, the steam is then expanded isentropically in a steam turbine so that the pressure returns to 0.7 bar and is cooled in a condenser. Determine:
a) Pump work
b) The incoming heat is given to the boiler
c) Turbine work
d) The heat removed by the condenser
e) Cycle thermal efficiency
Which correctly describes a different evolutionary stage of a star like the sun
A) it’s forms from a cold, dusty molecular cloud
B) During a yellow giant stage, it burns carbon in its core and helium in the shell surrounding the core.
C) After leaving the main sequence, its core is stable due to electron degeneracy
D) It becomes a white dwarf after exploding as a supernova
E)During a red giant stage, its core contracts and cools
The statement that correctly defines an evolutionary stage of a star like the sun is that after leaving the main sequence, its core is stable due to electron degeneracy. That is option C.
What are the stage of life cycle of a star?The stages of the life cycle of a star include the following:
Giant Gas CloudProtostarT-Tauri PhaseMain SequenceRed GiantThe Fusion of Heavier ElementsSupernovae and Planetary NebulaeThe evolutionary stage is also called the main sequence stage of the life cycle of the star.
In this stage, the core temperature reaches the point for the fusion to occur whereby the protons of hydrogen are converted into atoms of helium. This leads to the stability of the core of the newly formed start due to electron degeneracy.
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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.
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
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.
To learn more about Current, refer to:
Explanation:
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)
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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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
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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A soccer player kicks a ball of mass 0.500 kg toward the goal.The ball hits the crossbar at a height of 2.6 m with a speed of 15.0m/s. Suppose the ball was at rest on the ground before it was kicked. Use g = 9.80 m/s.
Answer:
The speed of the ball just before it hits the crossbar is 7.22 m/s.
Explanation:
We can use the conservation of energy to solve this problem.
At the moment the player kicks the ball, the ball has only kinetic energy, since it was at rest on the ground before being kicked. When the ball hits the crossbar, it has both kinetic energy and potential energy, since it is at a height above the ground. We can set the initial kinetic energy equal to the sum of the final kinetic and potential energy:
(1/2)mv^2 = mgh
where:
m = mass of the ball (0.500 kg)
v = initial speed of the ball (15.0 m/s)
g = acceleration due to gravity (9.80 m/s^2)
h = height of the crossbar above the ground (2.6 m)
We want to solve for the speed of the ball just before it hits the crossbar, which we can do by rearranging the equation:
v = sqrt(2gh)
v = sqrt(29.802.6) = 7.22 m/s (rounded to two decimal places
What are some examples of conservation of energy?
Answer:
power plant
collision
Battery
Burning wood
speaker
Beating drum
A stone is dropped in a mine shaft 15 m deep. The speed of sound is 343 m/s. How long does it take to hear the echo?
It takes 0.1311 seconds to hear the echo of the stone.
How to calculate the time it takes to hear the echo of the stone.First we need to determine the time it takes for the sound wave to travel from the stone to the bottom of the mine shaft and back up to our ears.
Let's start by finding the time it takes for the sound wave to reach the bottom of the mine shaft. We can use the formula:
time = distance / speed
The distance is the depth of the mine shaft, which is 15 meters. The speed of sound is 343 m/s, as given in the problem. Therefore, the time it takes for the sound wave to reach the bottom of the mine shaft is:
time = 15 m / 343 m/s
time = 0.0437 s
Now, we need to find the time it takes for the sound wave to travel back up to our ears. Since the sound wave travels at the same speed, 343 m/s, the distance it needs to cover is twice the depth of the mine shaft, or 30 meters. Therefore, the time it takes for the sound wave to travel back up to our ears is:
time = 30 m / 343 m/s
time = 0.0874 s
Finally, to find the total time it takes to hear the echo, we add the time it takes for the sound wave to reach the bottom of the mine shaft to the time it takes to travel back up to our ears:
total time = 0.0437 s + 0.0874 s
total time = 0.1311 s
Therefore, it takes 0.1311 seconds to hear the echo of the stone.
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The bigger the spring constant, the more__________the spring is.
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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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.
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.
HELP
Complete the ray diagram below:
The image characteristics are ____. (2 points)
A concave mirror is shown with curvature positioned at 8 on a ruler that goes from 0 to 14 centimeters. The object is located at 5, and the focal point is located at 6.5.
upright, virtual, and smaller
upright, real, and same size
inverted, virtual, and smaller
inverted, real, and same size
Real, inverted, and same size are the features of the image. when A concave mirror with a curvature of 8 is displayed on a ruler with a range of 0 to 14 cm.
The mirror formula may be used to calculate the image distance for an item located 4 cm from a 1.5 cm focal length mirror.
1/f = 1/u+1/v
f is the focal length
u is the object distance
v is the image distance
Keep in mind that the concave mirror's image distance and focal length are both positive.
Given:
u = 4cm
f = 1.5cm
1/v = 1/1.5-1/4
1/v = 0.67-0.25
1/v = 0.42
v = 1/0.42
v = 2.38cm
The picture is Genuine and INVERTED since the image distance value is positive.
We shall find its magnification and see if it is magnified or lessened. It is amplified if the magnification is larger than 1, and it is decreased if it is less.
Magnification = v/u
Magnification = 2.38/4
Magnification = 0.595 or. 0.6
The picture is reduced in size since the magnification is less than one (SMALLER).
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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.
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
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]
Pulse transfers a
disturbance. while wave is a
disturbance that transfers energy.
Answer:
Pulse transfers a single disturbance, while wave is a continuous disturbance that transfers energy.
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6. An 8000.0 kg truck starts off from rest and reaches a velocity of 18.0 m/s in 6.00 seconds. What is the truck’s acceleration and how much momentum does it have after it has reached this final velocity?
The truck's acceleration is 3.0m/s² and the momentum of the truck is 144000 kg m/s.
What is acceleration?It is the rate at which the speed and direction of a moving object vary over time.
We can use the following equation to calculate the acceleration of the truck:
a = (v - u) / t
where
a = acceleration
v = final velocity = 18.0 m/s
u = initial velocity = 0 m/s (the truck starts from rest)
t = time taken = 6.00 s
Substituting the values, we get:
a = (18.0 m/s - 0 m/s) / 6.00 s
a = 3.00 m/s²
Therefore, the acceleration of the truck is 3.00 m/s².
We can use the following equation to calculate the momentum of the truck:
p = m * v
where
p = momentum
m = mass of the truck = 8000.0 kg
v = final velocity = 18.0 m/s
Substituting the values, we get:
p = 8000.0 kg * 18.0 m/s
p = 144000 kg m/s
Therefore, the momentum of the truck after it has reached its final velocity is 144000 kg m/s.
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The driver of a car with a total of 1800 kg mass is traveling at 23 m/s when he slams on the brakes, locking the wheels on the dry pavement. The coefficient of kinetic friction between rubber and dry concrete is typically 0.7. How far would the car travel if were going twice as fast
Answer:
To solve this problem, we can use the formula:
d = (v^2)/(2μg)
d = distance traveled
v = speed of the car
μ = coefficient of kinetic friction
g = acceleration due to gravity
First, let's calculate the distance traveled when the car is traveling at 23 m/s:
d = (23^2)/(2*0.7*9.81) ≈ 67.97 meters
Now, let's calculate the distance traveled when the car is going twice as fast (46 m/s):
d = (46^2)/(2*0.7*9.81) ≈ 271.88 meters
Therefore, the car would travel approximately 271.88 meters if it were going twice as fast.
Sound travels through air at a speed of 342m/s
342
m
/
s
at room temperature. What is the frequency of a sound wave with a wavelength of 1.8m
1.8
m
Answer:
Explanation:
The formula relating the speed of sound, frequency, and wavelength is:
speed = frequency x wavelength
Rearranging this formula to solve for frequency:
frequency = speed / wavelength
Substituting the given values:
frequency = 342 m/s / 1.8 m
frequency = 190 Hz
Therefore, the frequency of the sound wave is 190 Hz.
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.
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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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.
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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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?
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.
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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
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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A 300 g football is kicked with an initial velocity of 140 m/s in a direction that
makes a 30° angle with the horizon. Find the peak height of the football.
Answer:
Explanation:
Assuming that air resistance is negligible, we can use the following kinematic equations to solve for the peak height:
v_f^2 = v_i^2 + 2ad
where v_f = 0 m/s (at the peak height) and a = -9.8 m/s^2 (acceleration due to gravity)
and
d = v_i t + (1/2)at^2
where d is the displacement or the peak height we want to find, v_i is the initial velocity, t is the time it takes to reach the peak height.
First, we need to resolve the initial velocity into its vertical and horizontal components:
v_i_x = v_i cos(30°) = 121.1 m/s
v_i_y = v_i sin(30°) = 70.0 m/s
Next, we can use the vertical component of the initial velocity to find the time it takes to reach the peak height:
v_f = v_i_y + at
0 m/s = 70.0 m/s + (-9.8 m/s^2)t
t = 7.14 s
Finally, we can use the time we found and the kinematic equation for displacement to find the peak height:
d = v_i_y t + (1/2)at^2
d = (70.0 m/s)(7.14 s) + (1/2)(-9.8 m/s^2)(7.14 s)^2
d = 247.5 m
Therefore, the peak height of the football is 247.5 meters.
A 208g sample of sodium-24 decays to 13.0g of sodium-24 within 60.0 hours. What is the half life of this radioactivity isotope?
Answer:
15 hours
Explanation:
formula: f(a) = a(0.5)^(T/t)
fill in known values: 13=208(0.5)^(60/t)
use natural log to isolate t: ln(13/208)=ln(0.5)(60/t)
solve for t: t=15
A student uses 800 W microwave for three seconds how much energy does a student use
Answer:
The student use 2400 Joules
Explanation:
From the formula E = pt
p = 800W
t = 3 seconds
=> E = 800*3 = 2400J
A 2.9 kg solid cylinder (radius = 0.20 m , length = 0.70 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.75 m high and 5.0 m long.
The final velocity of the cylinder is 1.22 m/s when it reaches the bottom of the ramp.
To solve this problem, we need to use conservation of energy and rotational kinematics.
Calculate the gravitational potential energy (GPE) of the cylinder at the top of the ramp:
GPE = mgh = (2.9 kg)(9.81)(0.75 m) = 21.39 J
Calculate the final kinetic energy (KE) of the cylinder when it reaches the bottom of the ramp:
[tex]KE = 1/2 mv^2 + 1/2 Iω^2[/tex]
where v is the linear velocity, I is the moment of inertia, and ω is the angular velocity.
Since the cylinder rolls without slipping, we know that v = ωr, where r is the radius of the cylinder.
[tex]KE = 1/2 mv^2 + 1/4 mv^2 = 3/4 mv^2 = 3/8 mgh[/tex]
Substituting the values we have:
KE = 3/8 (2.9 kg)(9.81)(0.75 m) = 63.56 J
Finally, we can use conservation of energy to find the final velocity of the cylinder:
GPE = KE
[tex]mgh = 3/8 mgh + 1/2 mv^2 + 1/2 Iω^2[/tex]
Solving for velocity:
[tex]v = \sqrt (2gh/5) = \sqrt(29.81 m/s^20.75 m/5) = 1.22 m/s[/tex]
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the complete question is:
At the top of a ramp, a 2.9 kg solid cylinder (radius = 0.20 m, length = 0.70 m) is released from rest and allowed to roll without slipping. The ramp measures 0.75 m in height and 5.0 m in length. calculate the final velocity when it reaches the bottom of the ramp
Select in the ticker-timer a frequency of 25 Hz or 50 Hz. Determine the period of the ticker-timer.
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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Match these items.
changes mechanical energy to heat energy, force x distance, rubbing energy ,using energy wisely, can cause heat pollution
work is done
collision
friction
stewardship
nuclear
energy
Mechanical energy to heat energy is collision,force x distance is work done,rubbing energy friction, stewardship is using energy wisely and nuclear energy can cause heat pollution.
EnergyThere are six different types of energy: chemical, electrical, radiant, mechanical, thermal, and nuclear. Other forms including electrochemical, auditory, electromagnetic, and others might be described in other study.Kinetic energy is the term for the energy that drives motion. Kinetic energy includes electrical and mechanical energy.Energy is the ability to conduct work in physics. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other other forms. Moreover, there is heat and work, which is energy moving from one body to another.For more information on energy kindly visit to
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What was the angle of application of the force of 35 if on a distance of 15 the work of 350 was done?
A current of O.S.A flows in a circuit with resistance 60 calculate the potential difference of the circuit
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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