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Leona [35]
2 years ago
7

A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displa

cement, the work it does is equal to how many joules?
Physics
2 answers:
pishuonlain [190]2 years ago
4 0
Data:
F = 20 [N]
m = 3 [kg]
s = 4 [m]
Work is calculated from the following formula:
W = F \cdot s \cdot cos(a)
Where a is the angle between force vector and displacement vector. When force is perpendicular to displacement, a = 90 deg and cosine of this angle = 0.
Therefore, the work this force does = 0 [J]
forsale [732]2 years ago
3 0
Work = (displacement) x (force in the direction of the displacement)

Since none of the force is in the direction of the displacement, the work it does
is zero. Something else is making the object move.  It's not the 20N force.
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A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting
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Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

\Delta U = n C_v(T_f-T_i)

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2    [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K

So,  \Delta U= 0.2\times12.5(1200-300)\\=2250 J

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J

Since, Q is positive 2250 J of heat will flow into the system.

6 0
2 years ago
in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at
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Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father
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6 0
2 years ago
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At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

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It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
2 years ago
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denis23 [38]
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