Learn Unit 3-Laws of Motion with Free Lessons & Tips

As a seasoned tutor on UrbanPro, I can guide you through this physics problem step by step.

Given: Initial velocity of the stone, u = 0 m/s (since it's dropped) Acceleration of the stone, a = g = 9.8 m/s^2 (acceleration due to gravity) Time, t = 11 s

(a) To find the velocity of the stone at t = 11 s, we can use the equation of motion:

v=u+atv=u+at

Substituting the values:

v=0+(9.8×11)v=0+(9.8×11) v=107.8 m/sv=107.8m/s

So, the velocity of the stone at t=11st=11s is 107.8 m/s107.8m/s.

(b) Now, to find the acceleration of the stone at t=11st=11s, we know that the acceleration due to gravity is acting on the stone throughout its motion, so the acceleration remains constant and equal to 9.8 m/s29.8m/s2.

Therefore, the acceleration of the stone at t=11st=11s is 9.8 m/s29.8m/s2.

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As an experienced tutor registered on UrbanPro, I'd be happy to help you with this physics problem.

When the string is cut, the bob will continue its motion with the velocity it had at the instant of cutting the string. Let's analyze both scenarios:

(a) When the bob is at one of its extreme positions: At the extreme position, all of the kinetic energy of the system is converted into potential energy. Therefore, the velocity of the bob at this point is zero. When the string is cut, the bob will simply fall vertically downward due to gravity.

(b) When the bob is at its mean position: At the mean position, the bob has its maximum kinetic energy and zero potential energy. The velocity of the bob at this point is given as 1 m/s. When the string is cut, the bob will continue its motion along a tangent to its trajectory at that point. Since the bob has no initial vertical velocity, it will follow a projectile motion under gravity.

So, in summary: (a) The trajectory is a simple vertical fall. (b) The trajectory is a projectile motion under gravity.

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Sure! Let's solve this problem step by step.

1. Identify the Forces: When the masses are released, the force due to gravity acts on each mass. The heavier mass (12 kg) will experience a greater force downward than the lighter mass (8 kg).

2. Apply Newton's Second Law: The net force on each mass will be the difference between the gravitational force pulling them down and the tension in the string pulling them up. The formula for this is:

   Fnet=m⋅aFnet=m⋅a

   where FnetFnet is the net force, mm is the mass, and aa is the acceleration.

3. Tension in the String: Since the string is light and inextensible, the tension in the string is the same throughout.

4. Acceleration Calculation: We can set up equations for each mass and solve them simultaneously.

   For the 8 kg mass: Fnet=T−mgFnet=T−mg 8a=T−8×9.88a=T−8×9.8 (as g=9.8 m/s2g=9.8m/s2)

   For the 12 kg mass: Fnet=mg−TFnet=mg−T 12a=12×9.8−T12a=12×9.8−T

5. Solve the Equations: Now, we can solve these equations simultaneously to find the acceleration and tension.

   Adding the two equations: 8a+12a=(12×9.8)−(8×9.8)8a+12a=(12×9.8)−(8×9.8) 20a=117.620a=117.6 a=117.620a=20117.6

   Now, plug the value of aa into one of the equations to find the tension:

   8(117.620)=T−8×9.88(20117.6)=T−8×9.8 T=8(117.620)+78.4T=8(20117.6)+78.4

6. Calculate: Let's calculate the values. a=117.620=5.88 m/s2a=20117.6=5.88m/s2 T=8(117.620)+78.4T=8(20117.6)+78.4

So, the acceleration of the masses is 5.88 m/s25.88m/s2, and the tension in the string is TT. You can plug in the value of TT to get the numerical value. This is how you can approach and solve this problem. If you need further assistance or clarification, feel free to ask!

Certainly! When tutoring on UrbanPro, I always strive to provide clear explanations to help students understand complex concepts like this one. Let's break down the scenario.

In the laboratory frame of reference, if a nucleus at rest disintegrates into two smaller nuclei, we need to consider the principles of conservation of momentum and conservation of energy.

1. Conservation of Momentum: Before the disintegration, the total momentum of the system (the nucleus at rest) is zero since the nucleus is stationary. After disintegration, the total momentum must still be zero according to the law of conservation of momentum. If the products move in opposite directions, their momenta will cancel each other out, resulting in a net momentum of zero.

2. Conservation of Energy: The total energy before and after the disintegration must remain constant. Since the nucleus was at rest initially, its energy is purely rest energy (mc^2). After disintegration, this energy is distributed among the kinetic energies of the two smaller nuclei. However, the total kinetic energy must still equal the initial rest energy to comply with the conservation of energy.

Therefore, for the two smaller nuclei to move in opposite directions, they must have equal but opposite momenta and equal kinetic energies. This ensures that both momentum and energy are conserved in the process.

In tutoring sessions on UrbanPro, I often illustrate these concepts with examples and diagrams to make them more accessible to students, helping them grasp the underlying physics principles effectively.

As an experienced tutor registered on UrbanPro, I'd be delighted to assist you with this physics problem.

When two billiard balls collide, they experience a change in momentum due to the impulse imparted during the collision. The impulse experienced by each ball can be calculated using the principle of conservation of momentum.

First, let's define the given values:

• Mass of each ball (m) = 0.05 kg
• Initial speed of each ball (u) = 6 m/s (opposite directions)
• Final speed of each ball after collision (v) = 6 m/s (rebounding with the same speed)

The change in momentum (∆p) of each ball can be calculated using the formula:

∆p = mv - mu

Since the final speed (v) is the same as the initial speed (u) for each ball, the change in momentum (∆p) simplifies to:

∆p = m(v - u)

Substituting the given values:

∆p = 0.05 kg * (6 m/s - (-6 m/s)) ∆p = 0.05 kg * (12 m/s) ∆p = 0.6 kg·m/s

Therefore, the impulse imparted to each ball due to the other is 0.6 kg·m/s. This implies that each ball experiences a change in momentum of 0.6 kg·m/s due to the collision with the other ball. If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is a great resource for finding online coaching and tuition services for physics and other subjects.

As an experienced tutor registered on UrbanPro, I can help you solve this physics problem. This question involves the concept of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided there are no external forces acting on the system.

Here, we have a shell fired by a gun. Let's denote the mass of the shell as msms and the mass of the gun as mgmg. The muzzle speed of the shell is vsvs, and we need to find the recoil speed of the gun, denoted as vgvg.

According to the conservation of momentum:

ms⋅vs=−(mg⋅vg)ms⋅vs=−(mg⋅vg)

Given:

• ms=0.020ms=0.020 kg
• mg=100mg=100 kg
• vs=80vs=80 m/s

Let's plug these values into the equation:

0.020×80=−(100×vg)0.020×80=−(100×vg)

1.6=−100vg1.6=−100vg

Now, let's solve for vgvg:

vg=1.6−100vg=−1001.6

vg=−0.016 m/svg=−0.016m/s

Now, since the negative sign indicates direction, the recoil speed of the gun is 0.016 m/s0.016m/s in the opposite direction of the shell's motion. This means the gun moves backward with a speed of 0.016 m/s0.016m/s when the shell is fired forward with a speed of 80 m/s80m/s.

So, the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N is approximately 17.32 m/s.

Feel free to ask if you have any questions or need further clarification!

As an experienced tutor registered on UrbanPro, I can certainly help with your physics question. The provided position-time graph of a particle of mass 0.04 kg depicts an interesting motion scenario. UrbanPro is indeed an excellent platform for online coaching and tuition needs, and I'm here to assist you.

To suggest a suitable physical context for this motion, let's analyze the graph. If we have a position-time graph where the particle undergoes repeated back-and-forth motion between two points, it could represent a scenario such as a simple harmonic oscillator. This could be exemplified by a mass attached to a spring oscillating in simple harmonic motion.

In this context, the time between two consecutive impulses received by the particle corresponds to the time taken for one complete oscillation cycle. This time period is commonly denoted as the period of oscillation, symbolized by 'T'.

To determine the magnitude of each impulse, we need to consider the change in momentum of the particle during each impulse. In simple harmonic motion, when the particle reaches the extremes of its motion (maximum displacement from equilibrium), it experiences maximum acceleration, hence maximum force. These extremes are the points where the impulses are applied, reversing the particle's direction.

Using the principles of impulse-momentum theorem, we can equate the impulse applied to the change in momentum of the particle. Since the momentum changes sign during each impulse, the magnitude of the impulse equals the change in momentum.

If we have access to additional information such as the amplitude of oscillation or maximum displacement, we can further refine our calculations. However, without such data, we might be limited to providing a general qualitative analysis.

In summary, the time between two consecutive impulses corresponds to the period of oscillation, and the magnitude of each impulse can be determined by analyzing the change in momentum of the particle at the extremes of its motion. If you have more specific data or further questions, feel free to share, and I'll be happy to assist you further.

As an experienced tutor registered on UrbanPro, I can help you tackle this physics problem effectively.

Firstly, let's address the net force acting on the man standing on the conveyor belt. When the belt is accelerating, two forces come into play: the force of friction between the belt and the man's shoes, and the man's weight.

1. Net Force on the Man: The net force is given by Newton's second law: Fnet=m⋅aFnet=m⋅a, where mm is the mass of the man and aa is the acceleration of the belt. In this case, the mass of the man is given as 65 kg and the acceleration of the belt is 1 m/s².

   So, Fnet=(65 kg)×(1 m/s2)=65 NFnet=(65kg)×(1m/s2)=65N.

2. Coefficient of Static Friction: The maximum frictional force that can exist between the man's shoes and the belt is determined by the coefficient of static friction, μsμs, and the normal force. The normal force in this case is equal to the man's weight, which is mgmg, where gg is the acceleration due to gravity (approximately 9.8 m/s29.8m/s2).

   The maximum static frictional force, Ffriction, maxFfriction, max, is given by μs×Normal Forceμs×Normal Force.

   So, Ffriction, max=μs×m×g=0.2×65×9.8=127.4 NFfriction, max=μs×m×g=0.2×65×9.8=127.4N.

To keep the man stationary relative to the belt, the force of static friction must balance the net force on the man. So, the maximum acceleration of the belt at which the man can remain stationary is the acceleration at which the force of friction equals the net force.

Ffriction, max=FnetFfriction, max=Fnet

⇒127.4 N=65 N⇒127.4N=65N

Therefore, the maximum acceleration of the belt that the man can withstand while remaining stationary relative to the belt is 1 m/s21m/s2. Beyond this acceleration, the man will start sliding on the belt due to the inability of static friction to counteract the increasing net force.