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Verdich [7]
4 years ago
12

In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9

.0 m above him, releases a ball from rest. While still in flight, the acrobat catches the ball just as it reaches him.Required:If he left the trampoline with a speed of 5.6 m/s, how long is he in the air before he catches the ball?
Physics
1 answer:
slega [8]4 years ago
6 0

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

t= 9/5.6

t= 1.6 secs

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Answer:

angle = 18.40 degree

Explanation:

given data

force = 95 N

distance = 0.50 m

torque = 15 N · m

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we will apply here torque equation that is express as

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15 = 0.50 × 95 × sin(θ)

sin(θ) = 0.315789

θ = 18.40 degree

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(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical
musickatia [10]

(a)

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Re-arranging it and replacing the numbers, we find the acceleration

a=\frac{v^2-u^2}{2S}=\frac{0-(25 m/s)^2}{2(120 m)}=-2.6 m/s^2

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(b)

In this case, the distance is different:

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3 years ago
What is Earth described as?
Ymorist [56]

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Explanation:

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