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Romashka-Z-Leto [24]
3 years ago
8

If the effort distance = 8 meters and the resistance distance = 1/2 meter. How much force does this man have to apply to lift th

e 1 ton (2000 N) weight?
Physics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

Explanation:

Effort x effort distance = load x resistance  distance

effort distance = 8 m ,

load = 2000N

resistance distance = 1/2 m = 0.5 m

Putting the values in the equation above

effort x 8m = 2000N x .5

effort = 2000 x 0.5 / 8

= 125 N

force required = 125 N .

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The speed of a 500 g cricket ball changes from 10m/s to 30m/s in just 7 seconds. What is the force acting on the cart?
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The force acting on the cart is 1.43 N.

<h3>What is force?</h3>

Force can be defined as the product of mass and acceleration.

To calculate the force acting on the cart, we use the formula below.

Formula:

  • F = m(v-u)/t................. Equation 1

Where:

  • F = Force acting on the cart
  • m = mass of the cart
  • v = Final velocity
  • u = initial velocity
  • t = time

From the question,

Given:

  • m = 500 g = 0.5 kg
  • v = 30 m/s
  • u = 10 m/s
  • t = 7 seconds

Substitute these values into equation 1

  • F = 0.5(30-10)/7
  • F = 10/7
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Hence, the force acting on the cart is 1.43 N.

Learn more about force here: brainly.com/question/13370981

7 0
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Discribe the law of reflection
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Answer:

the angle of reflection equals the angle of incidence—θr = θi. The angles are measured relative to the perpendicular to the surface at the point where the ray strikes the surface.

Explanation:

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y(t)=h/2     final height

So:

h/2=h-1/2*g*t^{2}

t=\sqrt{h/g}

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