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3 years ago

You dribble a basketball while walking on a basketball court. List and describe at least 3

Physics

2 answers:

Airida [17]3 years ago

4 0

Answer:

1. You push on the ball and the ball pushes on your hand .

2. The ball hits the ground and the ground pushes back on the ball .

3. You walk on the ground with your feet and the ground pushes back on you.

Explanation:

stellarik [79]3 years ago

4 0

Answer:

1. Walking

2. Pushing the basketball with hand

3. Ball bouncing off the ground

Explanation:

As per Newton's third law of motion every action has an equal and opposite reaction. Looking at the given scenario following are the pairs of action-reaction forces:

1. Walking: When you walk, you apply action and push the ground and the ground pushes you back with the same force.

2. Pushing the basketball with hand: when you push the ball that is action, the ball also applies force on your hand which is the reaction.

3. Ball bouncing off the ground: When ball hits the ground, it applies force i.e. action and the ground pushes it back with the same force which is the reaction.

You might be interested in

What electric force would a stationary 3.8 C charge experience if it were far away from any other charges

MAVERICK [17]

Answer:

The electric force will be 0 N

Explanation:

From the question we are told that

The magnitude of the charge is $q_1 = 3.8 \ C$

Generally from Coulombs law the electric force between two charges is mathematically represented as

$F = \frac{ k * q_1 q_2 }{r^2}$

Here r is the distance of separation between that two charges.

Now from the question we are told that the charge is far away from any other charge hence we can say that the distance between the charge and any other charge is $r = \infty$

$F = \frac{ k * 3.8 * q_2 }{\infty^2}$

=> $F =0 \ N$

Hence the electric force will be 0 N

3 0

3 years ago

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to

SpyIntel [72]

This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s

3 0

3 years ago

Circle the larger unit:

DerKrebs [107]

1. Centimeter
2. Kilogram
3. Millisecond
4. DL
5. Kg
6. Mm
7. S
8. Mm
9. Us

4 0

3 years ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

Part A: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

Part B: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$