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Free_Kalibri [48]
2 years ago
8

Sam is pulling a box up to the second story of his apartment via a string. The box weighs 53.3 kg and starts from rest on the gr

ound. Sam is pulling upwards with a force of 64 N. At the top, the box is pulled up to a height of 19.1 m. What is the total work performed for Sam to pull the box up to his height?
Physics
1 answer:
castortr0y [4]2 years ago
7 0

Answer:

W = 1222.4 J = 1.22 KJ

Explanation:

The work done on an object is the product of the force applied on it and the displacement it covers as a result of this force. It must be noted that the component of displacement in the direction of force should only be used. Hence, the work can be calculated as:

W = F d Cosθ

where,

W = Work Done = ?

F = Force Applied = 64 N

d = Distance Covered by Box = 19.1 m

θ = Angle between force and displacement = 0°

Therefore,

W = (64 N)(19.1 m)Cos 0°

<u>W = 1222.4 J = 1.22 KJ</u>

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Two resistors of 15 and 30 ω are connected in parallel. if the combination is connected in series with a 9. 0-v battery and a 20
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A light ray incident from medium 1 to medium 2, where n1&gt;n2. When the incident angle exceed the critical angle ac, the refrac
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Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}

Where,  

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{\theta_r} is the angle of refraction = 90°

{n_2} is the refractive index of the refraction medium

{n_1} is the refractive index of the incidence medium

Thus,

n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}

The formula for the calculation of critical angle is:

{sin\theta_{critical}}=\frac {n_2}{n_1}

Where,  

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3 years ago
Read 2 more answers
A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves wi
Vika [28.1K]

Answer:

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

Increase in internal energy Δu = nCVΔT

where n is the number of moles of the gas in vessel.

When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

1/2mv^2 = ∆u

1/2mv^2 = nCv∆T

Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

6 0
3 years ago
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