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Free_Kalibri [48]
4 years ago
8

After being struck by a bowling ball, a 1.7 kg bowling pin sliding to the right at 3.8 m/s collides head-on with another 1.7 kg

bowling pin initially at rest. Find the final velocity of the second pin in the following situations: a) The first pin moves to the right after the collision at 0.8 m/s. Answer in units of m/s.
Physics
1 answer:
Alchen [17]4 years ago
7 0

Answer:

3 m/s

Explanation:

Parameters given:

Mass of first bowling pin, m = 1.7 kg

Initial velocity of first bowling pin, u = 3.8 m/s

Final velocity of first bowling pin, v = 0.8 m/s

Mass of second bowling pin, M = 1.7 kg

Initial velocity of second bowling pin, U = 0 m/s

Let the final velocity of the second bowling pin be V

Using the principle of conservation of momentum:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1.7 * 3.8) + 0 = (1.7 * 0.8) + (1.7 * V)

6.46 = 1.36 + 1.7V

1.7V = 5.1

V = 5.1/1.7 = 3 m/s

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A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, i
Annette [7]

Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

AC^2=AB^2+BC^2

AC^2=(18.3)^3+(11.5)^2

AC=\sqrt{(18.3)^2+(11.5)^2}

AC=21.61\ km/s

Hence, The speed of meteoroid is 21.61 km/s in south-east.

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3 years ago
In this image of Earth and the Moon from the Sim, what does the circle around Earth represent?
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3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

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         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

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<u>Answer:</u>

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