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2 years ago

9

A 50 g tennis ball travels at a velocity of 15 m/s, hits a basketball with a mass of 600 g that is stationary on a frictionless

surface and then rebounds back in the opposite direction with a velocity of -6 m/s . How fast will the basketball be moving after the collision? m/s

1 answer:

statuscvo [17]2 years ago

8 0

Answer:

1.75 m/s

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(50 g) (15 m/s) + (600 g) (0 m/s) = (50 g) (-6 m/s) + (600 g) v

v = 1.75 m/s

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1. What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?

Anit [1.1K]

Answer:

<h3>The answer is 4.53 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

<h3>4.53 kgm/s</h3>

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2 years ago

3. In 1989, Michel Menin of France walked on a tightrope suspended under a

Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0

2 years ago

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

ankoles [38]

Answer:

The magnitude of the centripetal acceleration during the turn is $a=12.04\ m/s^2.$

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , $T=t\times 4=60\ s.$

Now , Velocity ,

$v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.$

Also , we know :

Centripetal acceleration ,

$a=\dfrac{v^2}{r}$

Putting all values we get :

$a=12.04\ m/s^2.$

Hence , this is the required solution .

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2 years ago

In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start

Hitman42 [59]

Answer:

$v_f = 16.6 m/s$

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

$F_g = mg sin\theta$

so the acceleration due to gravity along the plane is given as

$a = \frac{F_g}{m}$

now we have

$a = g sin\theta$

$a = (9.81 sin4.0)$

$a = 0.68 m/s^2$

now we know that

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$v_f = 16.6 m/s$

4 0

2 years ago

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic

larisa86 [58]

Answer:

The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

Explanation:

Given that,

Charge $q =8.4\times10^{-4}\ C$

Angle = 35°

Magnetic field strength $B=6.7\times10^{-3}\ T$

Magnetic force $F=3.5\times10^{-2}\ N$

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

$F = qvB\sin\theta$

$v = \dfrac{F}{qB\sin\theta}$

Where,

F = Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}$

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times0.57}$

$v = 10910.36\ m/s$

$v = 1.1\times10^{4}\ m/s$

Hence, The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

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2 years ago