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Law Incorporation [45]
3 years ago
14

Name the applied force in the free body diagram:

Physics
2 answers:
MAVERICK [17]3 years ago
6 0

The applied force is the force from an external agent that tries to start the motion or stop the motion. Since, the block is being dragged in the horizontal direction, the applied force will be spring balance force because it drags the block to the left. Hence, it is the applied force. The forces in the vertical direction are gravity and normal force and the dotted force is frictional force.

Answer: The spring scale drags the box.

guapka [62]3 years ago
5 0

The spring scale dragging the box, since applied force is a force acting on an object due to a person or another object.

Even though the table is an object, the table pushing up is the normal force, which is perpendicular to the contact force.

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Answer:

Kenetic

Explanation:

Potentiak is at the start

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An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found
Kaylis [27]

I assume the 100 N force is a pulling force directed up the incline.

The net forces on the block acting parallel and perpendicular to the incline are

∑ F[para] = 100 N - F[friction] = 0

∑ F[perp] = F[normal] - mg cos(30°) = 0

The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.

Then

F[friction] = 100 N

F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N

If µ is the coefficient of static friction, then

F[friction] = µ F[normal]

⇒   µ = (100 N) / (84.9 N) ≈ 1.2

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2 years ago
What are stars made from
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3 years ago
A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th
kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

P_1, V_1, T_1 & P_2, V_2, T_2

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as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

-Q=W+U_1----1

Q=-W+U_2-----2

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

U_1 & U_2 change in internal Energy of gas

Thus from 1 & 2 we can say that

U_1+U_2=0

n_1c_v(T-T_1)+n_2c_v(T-T_2)=0

T(n_1+n_2)=n_1T_1+n_2T_2

T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}

where n_1=\frac{P_1V_1}{RT_1}

n_2=\frac{P_2V_2}{RT_2}

T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}

T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}

and P=\frac{P_1V_1+P_2V_2}{V_1+V_2}

6 0
3 years ago
Acceleration is equal to the initial velocity minus the final velocity, then divided by time.true or false
tatiyna

Almost true but not quite.

That would give you the negative of the actual acceleration.

It should be the other way around:

(final v) minus (initial v), then divide by time.

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3 years ago
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