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tatyana61 [14]
3 years ago
13

When a particle of mass m is at (x comma 0 )​, it is attracted toward the origin with a force whose magnitude is StartFraction y

Over x squared EndFraction . If the particle starts from rest at x equals b and is acted upon by no other​ forces, find the work done on it by the time it reaches x equals a​, 0 less than a less than b.
Physics
1 answer:
Dahasolnce [82]3 years ago
8 0

Answer:

W = y (b-a) / ab

Explanation:

Work is defined by the expression

W = ∫ F. dr

In this case the force is in the same direction of displacement, so the scalar product is reduced to the ordinary product

W = ∫ F dr

The expression of the strength left is

F = -y / x²

let's replace and integrate

W = ∫ (-y / x²) dx

W = -y (-1 / x)

We evaluate between the lower limit x = b + a to the upper limit x = 0 + a

W = -y (-1 / b + 1 / a)

W = y (b-a) / ab

where (b-a) is the distance traveled

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Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th
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Answer:

A) G = m³/kg.s²

B) E = kg.m²/s²

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A)

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F = Gm₁m₂/r²

where, the units of all variables are:

F = Force = kg.m/s²

m₁ = m₂ = mass = kg

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Therefore,

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(kg.m/s²)(m²/kg²) = G

<u>G = m³/kg.s²</u>

<u></u>

B)

The given equation is:

E = mc²

where, the units of all variables are:

m = mass = kg

c = speed = m/s

E = Energy = ?

Therefore,

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This is the correct answer, which is not present in any option.

4 0
4 years ago
I need help on 11 - 14
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The answer is negative three

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3 years ago
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A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a
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Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination \theta = 15°

\sin \theta = \frac{AB}{AC}

\sin 15 = \frac{5900}{x}

x = \frac{5900}{0.2588}

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

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3 years ago
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