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Ket [755]
3 years ago
6

A rope horizontally pulls a massive object lying on a surface with friction with a constant

Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer:

Equal to the frictional force

Explanation:

<u>Question</u>; The options given with regards to a similar question posted online are;

A. Equal (equivalent) to the frictional force

B. Larger than the frictional force

C. Equal to the object's weight

D. More than the object's weight

Explanation

According to Newton's first law of motion, every object shall remain at rest or continue moving with uniform (constant speed) motion unless there is a net force acting on the object

Given that the velocity of the massive block, lying on the surface that has friction, being pulled by the rope = Constant

Therefore;

The net force acting on the moving block while being pulled by the rope = 0

From which we have;

The pulling force = The resistive force

Where;

The pulling force = The (pulling) force (applied) on the rope

The resistive force = The frictional force of the surface which tends to prevent the motion of the block

Therefore, given that the net force acting on the block = 0

The force on the rope = The frictional force (of the surface)

The correct option is option A. Equal to the frictional force.

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An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do
ipn [44]

Answer:

(a) ΔU=747J

(b) γ=1.3

Explanation:

For (a) change in internal energy

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Substitute the given values

ΔU=970J-223J

ΔU=747J

For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

Cv is the molar heat capacity at constant volume

To calculate γ we first need to find Cp and Cv

So

For Cp

As we know

Q=nCpΔT

Cp=(Q/nΔT)

C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\

So

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

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3 years ago
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Answer:

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Explanation:

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Answer:

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