Please need help with this

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball initially moving in th

inn [45]

Final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

<h3>What is conservation of momentum?</h3>

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. By the law of conservation of momentum,

$m_1u_1 + m_2u_2 = (m_1+m_2)v$

Here, (m) is the mass, (u) is initial velocity before collision, v is final velocity after collision and (subscript 1, and 2) are used for body 1 and 2 respectively. Rewrite the formula for final velocity as,

$v=\dfrac{m_1u_1 + m_2u_2}{(m_1+m_2)}$

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s, has a head-on collision with a 0. 090-kg ball, initially moving in the same direction at a speed of 3.44 m/s. Thus, the initial velocity of the second ball is,

$v_{2f}=5.82+3.44+v_{1f}\\v_{2f}=2.38+v_{1f}$

Let v1f is the final velocity of first ball. Thus, the initial velocity of the first ball is,

$v_{1f}=\dfrac{(0.060)(5.82) + (0.090)(3.44-2.38)}{(0.060)+(0.090)}\\v_{1f}=2.964\rm\; m/s$

Thus, final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

Learn more about the conservation of momentum here;

brainly.com/question/7538238

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4 0

2 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"

Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0

3 years ago

2.

saul85 [17]

Answer:

Explanation:

Planets have different year lengths because it depends how far they revolve from a celestial body. Each planet has its own orbital period. Planets closer to the star will have a lower orbital period compared to the ones that lie far away from it.

4 0

1 year ago

Read 2 more answers

(please help i gotta turn this in a few minutes 10 points!)

pogonyaev

Answer:

3a, 2b,4c,1d

Explanation:

what do I need to explain just something you know

7 0

3 years ago

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thr