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miskamm [114]
3 years ago
7

High speed stroboscopic photographs show that the head of a 244 g golf club is traveling at 57.6 m/s just before it strikes a 45

.2 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 39.9 m/s. Find the speed of the golf ball just after impact.
Physics
1 answer:
almond37 [142]3 years ago
6 0

Answer:

The speed will be "1.06 m/s".

Explanation:

The given values are:

Momentum,

m1 = 244 g

m2 = 45.2 g

On applying momentum conservation ,

Let v2 become the final golf's speed.  

From Momentum Conservation

⇒  Total \ initial \ momentum = Total \ final \ momentum

⇒  m1\times u1 + m2\times u2 = m1\times v1 + m2\times v2

On putting the estimated values, we get

⇒  0.244\times 57.6+0=0.244\times 39.9+45.2\times v2

⇒  57.844+0=9.7356+45.2\times v2

⇒  48.1084=45.2\times v2

⇒  v2=\frac{48.1084}{45.2}

⇒  v2=1.06 \ m/s

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If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
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The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0
3 years ago
A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h
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Answer:

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after 12 seconds, the radius becomes = \frac{dr}{dt} X 12 = 4 \frac{ft}{sec}  X 12 sec = 48 ft

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

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Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

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