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GaryK [48]
2 years ago
8

How far will a car travel if its velocity is 15 m/s in 3 seconds? Follow example below.

Physics
1 answer:
svet-max [94.6K]2 years ago
4 0
<h3><u>Solution</u><u>:</u></h3>
  • Distance (d) = 112 m
  • Time (t) = 4 seconds
  • Let the speed be v.
  • We know, speed = Distance / Time
  • Therefore, v = d/t

or, v = 112 m ÷ 4 s = 28 m/s

<h3><u>Answer</u><u>:</u></h3>

<u>The </u><u>speed </u><u>of </u><u>the</u><u> </u><u>cheetah</u><u> </u><u>is </u><u>2</u><u>8</u><u> </u><u>m/</u><u>s.</u>

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Find the hiker’s gravitational potential energy if the cliff is 60m high
Furkat [3]

Answer:

Potential energy is U=mgh

Explanation:

The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.

Potential energy  U=mgh

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g=9.81m/s^2

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U=m(9.81m/s^2)(60m)\\\\U=588.6*m

m is the mass of the hiker, wich is not in the description of the problem.

4 0
3 years ago
(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca
Aleks04 [339]

(a) Let v be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

a_{\rm rad} = \dfrac{v^2}R

where R is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

F = (1.500\,\mathrm{kg}) a_{\rm rad}

so that

(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N

Solve for v :

v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}

(b) The net force equation in part (a) leads us to the relation

F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}

so that v is directly proportional to the square root of R. As the radius R increases, the maximum linear speed v will also increase, so the cord is less likely to break if we keep up the same speed.

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1 year ago
A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
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W = 95.8J

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S = 6.00m

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