Answer:
We can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The heat transferred from the gold bar to the water is equal to the heat transferred from the water to the gold bar, since they reach thermal equilibrium. Therefore:
q_gold = q_water
We can solve for the temperature change of the gold bar:
q_gold = mcΔT_gold
q_water = mcΔT_water
Since the heat transferred is equal:
mcΔT_gold = mcΔT_water
Rearranging and solving for ΔT_gold:
ΔT_gold = ΔT_water(m_water/m_gold)
ΔT_water is the temperature change of the water, which is 17°C. m_water is 0.220 kg, and m_gold is 3 kg. c_gold is given as 129 J/kg K.
ΔT_gold = 17°C(0.220 kg/3 kg)(1/129 J/kg K) = 0.025°C
Therefore, the temperature change of the gold bar is 0.025°C when it is placed into 0.220 kg of water and thermal equilibrium is reached.
(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.
(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.
why the ocean near Christchurch is a different temperature than we’d expect for its latitude
Why the ocean near Christchurch is a different temperature than we'd expect for its latitude (distance from the equator)? Water moving from the equator is warmer than would be expected based on latitude, and so is warmer than the air it passes.
Changes to prevailing winds affect ocean currents. Changes to ocean currents affect how much energy is brought to (or taken away from) a location. In El Niño years, the prevailing winds that normally drive a warm current from the Equator past New Zealand are disrupted and may stop or even reverse.
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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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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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The units of voltage are
Oohms, amps, volts
Oamps, volts, ohms
Ovolts, ohms, amps
Ovolts, amps, ohms
of current are
and of resistance are
Answer:24 volts ÷ 4 amps = 6 ohms
Explanation:
We know that the Ohm's law has given the relationship between the current, voltage and the resistance of the wire. Mathematically, it can be written as :
Where
I is the current
R is the resistance
If current flowing is 4 amps and voltage is 24 volts. The formula to find the resistance will be :
R = 6 Ohms
Hence, the correct option is (d) " 24 volts ÷ 4 amps = 6 ohms ".
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
460miles per hour with the wind nd 420 per hour gainst the wind
The speed of the wind is 20 miles per hour.
To solve this problem, we can use the formula:
Speed = Distance/Time
Let's assume that the speed of the wind is x miles per hour.
With the wind, the plane travels at a speed of 460 miles per hour. This means that its speed relative to the ground is the sum of its airspeed and the speed of the wind:
460 = Airspeed + x
Against the wind, the plane travels at a speed of 420 miles per hour. This means that its speed relative to the ground is the difference between its airspeed and the speed of the wind:
420 = Airspeed - x
We can solve this system of equations to find the airspeed of the plane:
460 = Airspeed + x
420 = Airspeed - x
Adding the two equations gives:
880 = 2Airspeed
Dividing both sides by 2 gives:
Airspeed = 440 miles per hour
Now that we know the airspeed of the plane, we can find the speed of the wind by substituting this value into one of the equations we obtained earlier:
460 = Airspeed + x
460 = 440 + x
x = 20
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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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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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HELPPPP LATE HOMEWORK DUE TODAY
If F₁ has a greater magnitude than F₂, the box will accelerate backward because the net force is in the backward direction (1st option)
How do i know which direction the box will move to?To obtain the direction in which the box will move, we shall determine the net force acting on the box. This is illustrated below:
Assumption:
Magnitude of force 1 (F₁) = 40 NMagnitude of force 2 (F₂) = 25 NNet force (F) =?Net force = Magnitude of force 1 (F₁) - Magnitude of force 2 (F₂)
Net force = F₁ - F₂
Net force = 40 - 25
Net force = 15 N backward
From the above illustration, we can see that the net force is 15 N backward.
Thus, we can conclude from the box will accelerate backward (1st option)
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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.
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.
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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You leave Fort worth ,Texas,at 2:41 p.m. and arrive in Dallas at 3:23 p.m. , covering a distance of 58km. what is your average speed in metres per second ?
Answer:
Explanation:
The time taken to travel from Fort Worth to Dallas is:
t = 3:23 pm - 2:41 pm = 42 minutes = 0.7 hours
The distance covered is:
d = 58 km
The average speed is:
v = d/t = 58 km / 0.7 hours = 82.86 km/h
To convert km/h to m/s, we can use the conversion factor:
1 km/h = 0.2778 m/s
Therefore, the average speed in m/s is:
v = 82.86 km/h × 0.2778 m/s/km = 23.06 m/s (rounded to two decimal places)
So the average speed is 23.06 m/s.
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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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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A stuntman of mass 55 kg is to be launched horizontally out of a spring- loaded cannon. The spring that will launch the stuntman has a spring coefficient of 266N / m and is compressed 5 m prior to launching the stuntman. If friction and air resistance can be ignored, what will be the approximate velocity of the stuntman once he has left the cannon?
The approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.
StepsWe can use the conservation of energy, where the potential energy stored in the compressed spring is converted into the kinetic energy of the stuntman as he is launched out of the cannon.
The potential energy stored in the spring is given by:
PE = (1/2)kx²
where k is the spring constant and x is the distance the spring is compressed.
PE = (1/2)(266 N/m)(5 m)² = 3325 J
This potential energy is then converted into kinetic energy:
KE = (1/2)mv²
where m is the mass of the stuntman and v is his velocity.
3325 J = (1/2)(55 kg)v²
v² = (2*3325 J) / 55 kg
v² = 121 m²/s²
v = √(121 m²/s²) = 11 m/s
Therefore, the approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.
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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 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
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.
Projectile Motion Practice Problems (horizontal and at an angle)
1. Josh kicks a soccer ball with a velocity of 15 m/s at an angle of 38° above the
horizontal.
a. What are the X and Y components of his velocity?
b. How long is the ball in the air?
c. How far will the ball go?
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
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.
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Explanation:
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.
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.
(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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Gas pressure is defined as
Select one:
O a. mass per unit area
O b.
O c. force per unit area
O d. force per unit volume
acceleration per unit volume
Answer: b. force per unit area.
Explanation:
What was the angle of application of the force of 35 if on a distance of 15 the work of 350 was done?
As shown in the figure, a light inextensible string is passing around a smooth light pulley and it is attached to a spring having spring constant K. When a mass is suspended to the pulley, the pulley moved downwards through a distance of x and attained equilibrium. The mass attached to the pulley is given by :
1) Kx/g
2) 2Kx/g
3) 3Kx/g
4) 4Kx/g
5) 5Kx/g
Please show me how you worked it out, along with a brief explanation.
The mass attached to the pulley is Kx/g.
option 1.
What is the mass attached to the pulley?Assuming there is no friction, the tension in the string will be constant throughout, and the net force on the pulley-mass system will be the weight of the mass.
Let the mass of the hanging weight be m, then the weight of the mass is mg.
The spring will be extended by the same distance x, as the string is inextensible, and it will exert a force of Kx in the upward direction.
Since the pulley is in equilibrium, the net force on the pulley-mass system must be zero.
Therefore, the tension in the string must be equal to the force exerted by the spring:
T = Kx
Using the fact that the tension in the string is equal to the weight of the mass, we can write:
mg = Kx
Therefore, the mass attached to the pulley is:
m = Kx/g
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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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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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