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Sauron [17]
3 years ago
10

A 60 kg skateboarder accelerates themselves at 1.5m/s2. How much force was required to do this?

Physics
1 answer:
velikii [3]3 years ago
3 0

Answer:

90 N

Explanation:

m = 60 kg

a = 1.5 m/s^2

F = m × a = 1.5 × 60 = 90 N

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<h3>Following are the properties and characteristics of the glass.</h3>

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An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys
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Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

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input force×distance moved = output force × distance moved........................Equation 1

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
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Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

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<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

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\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

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It is the displacement of the particle in 2 seconds.

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