A wooden brick with mass M is suspended at the end of cords as shown above. A bullet with mass m is fired toward the brick with speed v0. The bullet collides with the brick embedding itself into the brick. The brick-bullet combination will swing upward after the collision. Consider the brick, earth, and bullet as part of a system. Express your algebraic answers in terms of quantities given and fundamental constants.
(a) During the collision of the brick and the bullet, compare the magnitude and direction of the impulse acting on the brick to the impulse acting on the bullet. Justify your answer.
(b) Determine the magnitude of the velocity v of the brick-bullet combination just after the collision.
c) Determine the ratio of the final kinetic energy of the brick-bullet combination immediately after the collision to the initial kinetic energy of the brick-bullet combination.
(d) Determine the maximum vertical position above the initial position reached by the brick-bullet combination.
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Answer: the answer given below
(a) Explanation: The impulse on an object is given by the change in momentum of the object. Before the collision, the bullet has momentum p1 = mv0 and the brick has momentum p2 = 0, since it is stationary. After the collision, the combined bullet-brick system has momentum p3.
Conservation of momentum requires that the total momentum before the collision is equal to the total momentum after the collision:
p1 + p2 = p3
mv0 + 0 = (m + M)V
where V is the velocity of the combined bullet-brick system after the collision. Solving for V, we get:
V = (mv0) / (m + M)
The impulse on the bullet during the collision is equal to the change in momentum of the bullet:
J_bullet = p3 - p1 = (m + M)V - mv0
Substituting the expression for V we found earlier:
J_bullet = (m + M)(mv0) / (m + M) - mv0 = 0
Therefore, the impulse on the bullet is zero during the collision.
On the other hand, the impulse on the brick during the collision is:
J_brick = p3 - p2 = (m + M)V - 0 = (m + M)(mv0) / (m + M) = mv0
Therefore, the magnitude of the impulse acting on the brick is equal to the initial momentum of the bullet, mv0, and it is in the same direction as the initial velocity of the bullet.
In summary, during the collision of the bullet and the brick, the impulse acting on the bullet is zero, while the impulse acting on the brick is mv0 in the direction of the initial velocity of the bullet.
(b) We can use the principle of conservation of momentum to solve for the velocity of the brick-bullet combination just after the collision. The total momentum of the system (bullet, brick, and Earth) is conserved before and after the collision. Initially, only the bullet has momentum, which is given by p1 = m*v0, and the momentum of the brick and Earth is zero. After the collision, the bullet becomes embedded in the brick, and the combined system of the brick-bullet has momentum p2. Since the momentum of the Earth is negligible compared to that of the bullet and brick, we can treat the system as closed and apply conservation of momentum:
p1 = p2
m*v0 = (M + m)*v
where v is the velocity of the combined system just after the collision.
Solving for v, we get:
v = (m*v0) / (M + m)
Therefore, the magnitude of the velocity of the brick-bullet combination just after the collision is:
|v| = |(m*v0) / (M + m)|
The direction of the velocity is upward, as the system swings up after the collision due to the conservation of momentum.
(c) The initial kinetic energy of the system is the kinetic energy of the bullet just before the collision, which is given by:
KE1 = (1/2)mv0^2
The final kinetic energy of the system is the kinetic energy of the combined brick-bullet system just after the collision, which is given by:
KE2 = (1/2)*(M + m)*v^2
Substituting the expression we found for v:
KE2 = (1/2)(M + m)[(mv0) / (M + m)]^2
KE2 = (1/2)(m*v0^2) / (1 + M/m)
The ratio of the final kinetic energy to the initial kinetic energy is:
KE2 / KE1 = [(1/2)(mv0^2) / (1 + M/m)] / [(1/2)mv0^2]
KE2 / KE1 = 1 / (1 + M/m)
Therefore, the ratio of the final kinetic energy of the brick-bullet combination immediately after the collision to the initial kinetic energy of the brick-bullet combination is:
KE2 / KE1 = 1 / (1 + M/m)
(d)To determine the maximum vertical position reached by the brick-bullet combination, we can use conservation of energy, assuming there is no energy loss due to friction or other dissipative forces. At the maximum height, the kinetic energy of the system is zero, and all the initial kinetic energy has been converted to potential energy due to the height above the initial position.
The initial total energy of the system is the sum of the initial kinetic energy of the bullet and the gravitational potential energy of the brick:
E1 = (1/2)mv0^2 + Mgh1
where h1 is the initial height of the brick above the ground, and g is the acceleration due to gravity.
At the maximum height, the final total energy of the system is the potential energy due to the height above the ground:
E2 = (M + m)gh2
where h2 is the maximum height reached by the brick-bullet combination above the initial position.
Since there is no energy loss, we can set the initial energy equal to the final energy:
E1 = E2
Substituting the expressions for E1 and E2 and solving for h2, we get:
(M + m)gh2 = (1/2)mv0^2 + Mgh1
h2 = [(1/2)mv0^2 + Mgh1] / [(M + m)*g]
Simplifying, we get:
h2 = (1/2)v0^2 / g + h1(M/m) / (1 + M/m)
Therefore, the maximum vertical position above the initial position reached by the brick-bullet combination is:
h2 = (1/2)v0^2 / g + h1(M/m) / (1 + M/m)
Hope this helps :)
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The amount of force that the pitcher applies to the baseball is 11.6N.
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According to this question, a baseball has a mass of 145 g. A pitcher throws the baseball so that it accelerates at a rate of 80 m/s². The force applied on the baseball can be calculated as follows:
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An empty beer can has a mass of 50 g, a length of 12 cm, and a radius of 3.3 cm. Assume that the shell of the can is a perfect cylinder of uniform density and thickness.
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(b) What is the mass of the shell?
(c) Find the moment of inertia of the can about the cylinder's axis of symmetry.
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Empty beer can: mass 50g, length 12cm, radius 3.3cm. Moment of inertia found by subtracting mass of lid/bottom from mass of empty can, and using I=(1/2)mr² for a solid cylinder. Result: 1.7 x 10^-5 kg m².
An empty beer can has a mass of 50 g, a length of 12 cm, and a radius of 3.3 cm. Assume that the shell of the can is a perfect cylinder of uniform density and thickness. To find the moment of inertia of the can about the cylinder's axis of symmetry-
(a) Let the mass of the lid/bottom be m. The mass of the empty can is 50g.
Since the lid and bottom are identical in shape and mass, we can write that the total mass of the can is 2m + 50g.
Thus, the mass of the lid/bottom is m = (50g)/2 = 25g.
Therefore, the mass of the lid/bottom is 25g.
(b) The mass of the shell is the mass of the empty can minus the mass of the lid/bottom.
Therefore, the mass of the shell is
[tex]m_{shell} = m_{empty} - m_{lid/bottom} = 50g - 25g = 25g.[/tex]
(c) Moment of inertia of a solid cylinder of radius r and mass m about the axis of symmetry is given by
I = (1/2)mr²
The radius of the can is r = 3.3 cm = 0.033 m.
The length of the can is not needed to find the moment of inertia of the can about its axis of symmetry since the moment of inertia is independent of the length of the cylinder (as long as its mass and radius remain the same).
The mass of the shell is m_shell = 25g = 0.025 kg.
Using the formula for moment of inertia, we get
[tex]I = (1/2)mr² = (1/2)(0.025 kg)(0.033 m)² = 1.7 x 10^-5 kg m²[/tex]
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As a result, when the magnetized bar is brought near the other bar, it will stick to it, indicating that it is magnetized. The bar that does not stick is unmagnetized. Thus, with the aid of two bars, one magnetized and the other unmagnetized, we can determine which is which.
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If a 21.00-kg child slide from a height of 3.40 m above the bottom of the slide and her speed at the bottom is 2.30 m/s, the amount of energy lost due to friction is 644.18 J.
The potentiаl energy of аn object depends on the locаtion of the object from the bottom reference floor аnd the mаss of the object. The аmount of energy contаins by the object аt аny height is known аs the potentiаl energy of thаt object.
We are given:
The energy of the child at the upper end of the slide is,
[tex]E_{u}[/tex] = mgh
Substitute the values in the above equation
[tex]E_{u}[/tex] = 21 kg × 9.8 m/s2 × 3.40 m
= 699.72 J
The energy at the bottom of the slide is,
[tex]E_{b}[/tex] = [tex]\frac{1}{2}(mv^{2})[/tex]
Substitute the values in the above equation.
[tex]E_{b}[/tex] = [tex]\frac{1}{2}(21.2.30^{2})[/tex]
[tex]E_{b}[/tex] = 55.54 J
The energy lost due to friction is,
[tex]E_{f}[/tex] = [tex]E_{u}[/tex] - [tex]E_{b}[/tex]
Substitute the values in the above equation
[tex]E_{f}[/tex] = 699.72 - 55.54
[tex]E_{f}[/tex] = 644.18 J
Thus, the energy lost due to friction is 644.18 J.
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The camera should be now focused at a distance of 16 meters.
The camera, in this case, should focus on the distance from the mirror to the object reflected by the mirror. The distance should be twice the distance of the object to the mirror.
The mirror image and the object should be equidistant from the mirror. This implies that the distance of the object from the mirror is equal to the distance of the mirror image from the mirror.
The distance that the camera should focus on is equal to the distance from the object to the mirror, multiplied by 2. Therefore, Distance from the object to the mirror = 8 meters
Distance from the camera to the object = distance from the mirror to the object, which is twice the distance from the mirror to the object
Distance from the camera to the object = 2 × 8 meters = 16 meters
Therefore, the camera should be focused at a distance of 16 meters.
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Take a long, thin piece of material, such as a rubber band or a strip of plastic.2. Stretch it taut between two points, such as two pencils or two pegs.3. Pluck the string with your finger and observe the vibration.4. Vary the tension and length of the string to produce different pitches.
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what is the speed acquired by a freely falling object 5 s after being dropped from a rest position? what is the speed 6 s after?
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The speed acquired by the body is 49m/s and 59m/s respectively.
The speed can be calculated using the formula:
v= u + gt, where v= final speed, u= initial speed = 0 for a freely falling body, g= acceleration due to gravity, t= time.
The speed acquired by a freely falling object 5 seconds after being dropped from a rest position is 49 m/s. This is because an object dropped from rest will accelerate at a rate of 9.8 m/s², so after 5 seconds it will be moving at a speed of 5 * 9.8 = 49 m/s.
The speed 6 seconds after being dropped from a rest position is approximately 59 m/s. This is because an object dropped from rest will accelerate at a rate of 9.8 m/s², so after 6 seconds it will be moving at a speed of 6 * 9.8 = 58.8 m/s.
In summary, the speed of an object dropped from rest 5 seconds after being dropped is 49 m/s, and 6 seconds after it is approximately 59 m/s.
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The average speed of the particle is 4.7 calculated by dividing the total distance traveled by the time taken.
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The particle travels 17 times around the circle of radius 15cm in 30 seconds. Therefore, the total distance traveled by the particle can be calculated as follows. Total Distance = 17 × Circumference. Total Distance = 17 × 94.25cm. Total Distance = 1602.25cm. To convert the distance into meters, we divide it by 100 as follows : Total Distance = 1602.25cm = 16.0225m. Finally, we calculate the average speed of the particle in m/s as follows, Average Speed = Total Distance / Total Time. Average Speed = 16.0225m / 30s. Average Speed = 0.534m/s × 8.75 = 4.7. Therefore, the particle's average speed in m/s is 4.7.
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The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is 100 km/h. The air is moving in the wind at 31.0 km/h toward the north. The velocity of the airplane relative to the ground is: 104 km/h
The airplane's velocity relative to the ground is calculated by adding the velocity of the airplane relative to the air with the velocity of the air relative to the ground.
The velocity of the airplane relative to the ground is obtained by vector addition of the airplane's velocity relative to the air and the air's velocity relative to the ground. Given that the compass indicates a heading due west, the airplane's velocity relative to the air is 100 km/h towards the west.
The air is moving towards the north at 31.0 km/h, therefore the velocity of the air relative to the ground will be towards the north. The velocity of the air relative to the ground will be equal to 31.0 km/h towards the north.
To find the velocity of the airplane relative to the ground, we need to add the velocity of the airplane relative to the air to the velocity of the air relative to the ground.
Hence, we get the velocity of the airplane relative to ground = velocity of the airplane relative to air + velocity of air relative to ground. The velocity of the airplane relative to the ground = (100 km/h)2 + (31.0 km/h)2 = 104 km/h.
The velocity of the airplane relative to the ground is 104 km/h.
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Answer : If an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2 then after 5 seconds, the object will have a speed of 100 m/s
This can be calculated using the equation v = a*t, where v is the velocity, a is the acceleration due to gravity, and t is the time elapsed. Therefore, in this case, v = 20 m/s2 * 5 s = 100 m/s. These values are given in question, so we just have to put them in equation.
Since the object is falling freely, its acceleration remains constant and it follows a uniform acceleration motion. Therefore, the velocity of the object will increase linearly with time. After 10 seconds, the velocity will double to 200 m/s, and so on.
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if we say that the potential at the earth's surface is 0 v , what is the potential 1.6 km above the surface?
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If we say that the potential at the earth's surface is 0 v , the potential 1.6 km above the surface is - 6.2 × 10^6 V.
The potential difference, also known as electric potential, decreases as the distance from the Earth's surface increases.
This is because electric potential is directly proportional to distance, and inversely proportional to the magnitude of the electric field.
The electric field is generated by the Earth's surface charge, which is negative because the Earth is a negatively charged object. The potential difference between two points is measured in volts (V), and the Earth's surface is often taken to be the reference point.
If the potential at the Earth's surface is taken to be 0 V, the potential 1.6 km above the surface can be calculated as follows:
The electric field generated by the Earth's surface charge is given by: E = kq/r²,
where k is Coulomb's constant, q is the surface charge of the Earth, and r is the distance from the center of the Earth.
The potential difference between two points is given by: V = Ed,
where d is the distance between the two points.
Thus, the potential at a point 1.6 km above the Earth's surface is:
V = E × d = kq/r² × d = (9 × 10^9 N·m²/C²) × (- 5.52 × 10^5 C)/[(6.38 × 10^6 m + 1.6 × 10^3 m)²] × (1.6 × 10^3 m)
= - 6.2 × 10^6 V.
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a wrench is used to tighten a nut. a 15n perpendicular force is applied 50cm away from the axis of rotation, and moves a distance of 10 cm as it turns. what is the torque applied to the wrench?
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The torque applied to the wrench can be calculated using the formula:
torque = force x distance
where force is the perpendicular force applied, and distance is the distance from the axis of rotation at which the force is applied.
So, torque = 15 N x 0.5 m = 7.5 Nm
However, since the force moves a distance of 10 cm as it turns, the work done is:
work = force x distance moved = 15 N x 0.1 m = 1.5 J
This means that some of the energy applied by the force is lost to friction or other factors, and not all of it is converted into torque.
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Using this circuit below, find the Norton's equivalent circuit about terminals a and b. Req and leg are the equivalent resistance and current used in the Norton's equivalent ciruict. V1 = 10 V, R1 = 4ohms, R2 = 8ohms „R₃ = 8ohms Select one: a. leq = -2.5 A, Req = 2 ohms b. leq = 2.5 A, Req = 2 ohms c. leq = 2.5 A, Req = 64 ohms d. leq = -2.5 A, Req = 12.8 ohms
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The Norton's equivalent circuit and equivalent resistance of the given circuit is leq = 2.5 A, Req = 2 ohms. The correct answer is option b.
Norton's equivalent current, iNorton is calculated by dividing the voltage source by the series resistance of R2 and R3.
iNorton = V1 / (R2 + R3)
iNorton = 10 / (8 + 8)
iNorton = 0.625 A
Norton's equivalent resistance, RNorton is calculated by using the formula;
RNorton = R2 || R3
RNorton = (R2 x R3) / (R2 + R3)
RNorton = (8 x 8) / (8 + 8)RNorton = 4 ohms
Therefore, Norton's equivalent circuit is given by the current source of 0.625 A and the resistance of 4 ohms, connected across terminals a and b. The correct answer is option B; leq = 2.5 A, Req = 2 ohms.
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what is the distance between your eye and the image of the butterfly in the mirror? explain your answer.
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The distance between your eye and the image of the butterfly in the mirror is: the same as the distance between your eye and the actual butterfly
The distance between your eye and the image of the butterfly in the mirror is the same as the distance between your eye and the actual butterfly, which is the sum of the distance from your eye to the mirror and the distance from the mirror to the butterfly.
To calculate this, we need to measure the distance from your eye to the mirror, which can be done using a ruler or tape measure, and then measure the distance from the mirror to the butterfly, which can be done using a ruler or tape measure as well. Once we have these two measurements, we can simply add them together to get the total distance between your eye and the image of the butterfly in the mirror.
To clarify further, let's use an example. If your eye is 10 cm away from the mirror and the butterfly is 30 cm away from the mirror, then the total distance between your eye and the image of the butterfly in the mirror is 40 cm. This is because 10 cm (from your eye to the mirror) + 30 cm (from the mirror to the butterfly) = 40 cm.
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a student exerts a horizontal force of 40.0 n with her hand and pushes a 10.0 kg box a distance of 2.0 m across a frictionless floor. calculate the magnitude of the work done by the student. group of answer choices 40.0 j 60.0 j 80.0 j 100.0 j
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The magnitude of the work done by the student is 80.0 J. Option c is correct.
The work done by the student can be calculated using the formula,
W = Fd cos(theta)
where W is the work done, F is the force exerted, d is the distance moved, and theta is the angle between the force vector and the displacement vector.
In this problem, the force exerted by the student is a horizontal force of 40.0 N, and the box is moved a distance of 2.0 m across a frictionless floor. Since the force and displacement vectors are in the same direction (horizontal), the angle between them is 0 degrees, so cos(theta) = 1. Therefore, we can calculate the work done as,
W = (40.0 N)(2.0 m) cos(0) = 80.0 J
Hence, option c is correct choice.
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if 22.5L of nitrogen at 748 mm Hg are compressed to 725 mm hg at constant temperature what is the new volume?
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The new volume is approximately 23.16 L when the nitrogen gas is compressed from 748 mmHg to 725 mmHg at constant temperature.
Use the combined gas law to determine the relationship between a gas's pressure, volume, and temperature:
P1V1/T1 = P2V2/T2
where the gas's starting pressure, volume, and temperature are P1, V1, and T1, and its ultimate pressure, volume, and temperature are P2, V2, and T2.
The equation may be made simpler by saying: since the temperature is constant.
P1V1 = P2V2
Substituting the given values, we get:
725 mmHg × V2 = 748 mmHg × 22.5 L
Solving for V2, we get:
V2 = (748 mmHg × 22.5 L) / 725 mmHg
V2 = 23.16 L
A gas law known as the combined gas law connects a gas's pressure (P), volume (V), and temperature (T). It combines Boyle's law, Charles' law, and Gay-law, Lussac's three additional gas laws.
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What causes an object to become electrically charged?
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An object becomes electrically charged when there is a transfer of electrons between two objects. Electrons are negatively charged particles that orbit the nucleus of an atom. When two objects come into contact with each other, some electrons may move from one object to the other. The object that loses electrons becomes positively charged, while the object that gains electrons becomes negatively charged.
This transfer of electrons can also occur without direct contact between the objects. For example, if a charged object is brought close to a neutral object, the electrons in the neutral object may be attracted or repelled by the charged object. This can cause the electrons in the neutral object to move around, resulting in a separation of charges and the object becoming charged.
Another way an object can become charged is through the process of induction. If a charged object is brought near a neutral object, it can induce a separation of charges in the neutral object. This happens because the charged object creates an electric field that attracts or repels electrons in the neutral object. The result is a separation of charges, with one part of the object becoming positively charged and the other part becoming negatively charged.
what is the potential difference between two points in an electric field if 1 j of work is required to move 1 c of charge between the points
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The potential difference between the two points in an electric field is 1 V.
Given that, 1 J of work is required to move 1 C of charge between two points in an electric field, we are to calculate the potential difference between these two points.
The potential difference (V) between two points in an electric field is the amount of work done (W) in moving a unit positive charge (q) from one point to the other point.
Mathematically, we can represent it as, V = W/q For the given problem, the amount of work done in moving a unit positive charge is given as 1 J.
So we can write it as, W = 1 J Also, the amount of charge moved is 1 C. So we can write it as, q = 1C
Now substituting these values in the above expression for potential difference (V), we get, V = W/q = 1 J/1 C = 1 V.
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suppose you were dragging a table across a rough floor. in this case, the potential energy for friction depends on which quantity or quantities? (choose all that apply)
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In dragging a table across a rough floor, the potential energy for friction depends on the coefficient of friction, normal force, and distance traveled by the table, hence option (a), (b), and (c) are correct.
In this case, the potential energy for friction would depend on the following quantities:
Coefficient of friction: The coefficient of friction between the table and the floor would determine how much force is required to move the table and hence, the potential energy for friction.
Normal force: The normal force acting on the table due to the weight of the table and any objects placed on it would also affect the potential energy for friction.
Distance moved: The distance the table is moved would determine the amount of work done against friction and hence, the potential energy for friction.
Surface area: The surface area in contact between the table and the floor could also affect the potential energy for friction.
Overall, the potential energy for friction depends on a combination of factors, including the properties of the surfaces in contact, the force required to move the object, and the distance moved.
Therefore correct options are (a), (b), and (c).
Suppose you were dragging a table across a rough floor. in this case, the potential energy for friction depends on which quantity or quantities? (choose all that apply)
a. The total distance the table travels.
c. The coefficient of friction between the table and the floor.
d. The normal force that the floor exerts on the table.
e. There is no potential energy for frictional forces.
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the paper dielectric in a paper-and-foil capacitor is 8.10*10^-2 mm thick. it's dielectric constant is 2.10, and it's dielectric strength is 50.0 MV/m. assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Part A: What area of each plate is required for for a 0.300 uF capacitor? In m^2
Part B: If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor? In V
Answers
a. Part A: The area of each plate is required for for a 0.300 uF capacitor is 1.56 × [tex]10^{-4}[/tex] m².
b. Part B: If the electric field in the paper is not to exceed one-half the dielectric strength, the maximum potential difference that can be applied across the compactor is 2025 V.
To find the area of each plate required for a 0.300 uF capacitor, use the formula:
C = ε₀εrA/d
where C is the capacitance, ε₀ is the vacuum permittivity (8.85 × [tex]10^{-12}[/tex] F/m), εr is the relative permittivity (dielectric constant), A is the area, and d is the distance between the plates. In this case,
C = 0.300 uF
εr = 2.10
d = 8.10 × [tex]10^{-5}[/tex] m.
Rearrange the formula to find A:
A = Cd / (ε₀εr)
A = (0.300 × [tex]10^{-6}[/tex] F)(8.10 × [tex]10^{-5}[/tex] m) / (8.85 × [tex]10^{-12}[/tex] F/m × 2.10)
A ≈ 1.56 × [tex]10^{-4}[/tex] m²
Thus, the area of each plate required for a 0.300 uF capacitor is approximately 1.56 × [tex]10^{-4}[/tex] m².
To find the maximum potential difference that can be applied across the capacitor, use the formula:
V = Ed
where E is the electric field and d is the distance between the plates. In this case, E is half the dielectric strength (50.0 MV/m / 2 = 25.0 MV/m), and d = 8.10 × [tex]10^{-5}[/tex] m:
V = (25.0 × 10^6 V/m)(8.10 × 10^-5 m)
V ≈ 2025 V
Thus, the maximum potential difference that can be applied across the capacitor without exceeding one-half the dielectric strength is approximately 2025 V.
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