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Gwar [14]
2 years ago
10

A hockey player makes a slap shot exerting a constant force of 40.0 newtons on the puck for .2 seconds. What is the magnitude of

the impulse given to the puck?
Physics
1 answer:
Taya2010 [7]2 years ago
6 0

Answer:

Magnitude of impulsive force is <u>8 N.s</u>.

Explanation:

Given:

Force acting on the puck is, F=40.0\ N

Time interval for which the force acts is, \Delta t=0.2\ s

We are asked to find the impulsive force.

Impulsive force acting on a body is the sudden change in the momentum of the body due to the application of a large constant force for a very small interval of time.

Here, the hockey player applies a large force of 40 N for a very short interval of time. So, the impact on the puck is measured as Impulse and is represented by 'J'.

Impulse in terms of applied force and time interval is given as:

J=F\Delta t

Plug in the given values and solve for 'J'. This gives,

J=40\times 0.2\\J=8\ N\cdot s

Therefore, the magnitude of impulsive force is 8 N.s.

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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

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Thus,

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We are told the salmon reached a height of 0.294 m

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g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

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v_y = vsinθ

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0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

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t = √2.885/4.905

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Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

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For the answer to the question above, asking what is the resistance force if a man weighing 700 N lifts a 50.0 N object 2.0 m high. He uses a lever and with a force of 10.0 N. 
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