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Mice21 [21]
3 years ago
11

Part a. A 10 Kg object is pulled by a force of 60 N at an angle of 30°. As it moves on, it is resisted by a force of friction of

2 N, as shown by the diagram below. Determine the normal force F normal
part b. A 10 Kg object is pulled by a force of 60 N at an angle of 30°. As it moves on, it is resisted by a force of friction of 2 N, as shown by the diagram below. determine the acceleration.

Physics
1 answer:
Tom [10]3 years ago
6 0

Answer:

What cell organelle makes turgor pressure in a plant cell

possible? Describe the role of this organelle in this process.

Explanation:

pls help dude

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when a book falls from a shelf to the ground without anyone pushing on it, no work is done to the book. True or False
VikaD [51]

That's false.

-- The force on the book is its weight.

-- The force acts through the distance the book falls.

-- The work done on the book is

(weight) x (length of the fall)

-- Gravity DOES the work.

8 0
3 years ago
Find the gravitational potential energy of a light that has a mass of 13.0 kg and is 4.8 m above the ground
Lynna [10]

Answer:

611.52 <---- Is your answer

Explanation:

If you need explanation mark me as brain list

8 0
3 years ago
You use a pulley system to lift a car engine. You apply a force of 120n and the pulley pulls on the engine with a force of 1050n
Savatey [412]

Given,

Effort force = 120 N

Load force= 1050 N

Mechanical advantage of a pulley is given by the ratio of load force to the effort force.

M.A=\frac{Load force}{Effort force}

=\frac{1050}{120}

=8.75

Therefore, the mechanical advantage of the given pulley is 8.75.

5 0
3 years ago
Read 2 more answers
Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion
stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

= initial volume of gas  =  

= final volume of gas  =  

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0
3 years ago
Numeria
Vinil7 [7]

Answer:

hope it helps

plz mark as brainliest

5 0
3 years ago
Read 2 more answers
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