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4 years ago

A mule pulls a cart of milk 10 meters with a force of 50 Newton's.Calculate the work done by a mule

Physics

2 answers:

san4es73 [151]4 years ago

8 0

Work = Force x Distance

Assuming that this work is being done parallel to the displacement that is, but under that assumption:

W = (50)(10)

W = 500 J

Alex73 [517]4 years ago

4 0

Answer: W = (50)(10)

Explanation:

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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

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4 years ago

Need help with these 2 questions ??????!!!!! Hurry this is due soon

Readme [11.4K]

Answer: The first one is A and C

Explanation:

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3 years ago

Read 2 more answers

What is the fundamental cause of air circulation in earths atmosphere

Digiron [165]

<span>A differences in the warmth, or moisture level as well as neighbouring areas of pressure in </span>air cause air<span> to circulate. in the earth's atmosphere.</span>

4 0

3 years ago

Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs.

Step2247 [10]

Answer:

Part a)

$F = 15600 N$

Part b)

force is 21.2 times more than the weight of the person

Explanation:

Part a)

As it is given that the distance after which he stopped is given as

d = 1.50 cm

here finally it stops so final speed is given as

$v_f = 0$

initial speed is given as

$v_i = 6 m/s$

now by the equation of kinematics we know that

$v_f^2 - v_i^2 = 2a d$

$0^2 - 6^2 = 2(a)(0.015)$

$a = -1200 m/s^2$

Now the force on the leg is given as

$F = ma$

m = mass of leg = 13 kg

$F = (13 kg)(-1200 m/s^2)$

$F = 15600 N$

Part b)

Force due to weight of the object

$F_g = Mg$

here we know

M = 75.0 kg

$F_g = 75(9.81)$

$F_g = 735.75 N$

Now we know that the ratio of the weight with the force on leg is given as

$\frac{F}{F_g} = \frac{15600}{735.75} = 21.2 N$

so force is 21.2 times more than the weight of the person

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4 years ago

what would happen to the surface temperature on Earth if large amounts of carbon dioxide were removed from the atmosphere?

slava [35]

Answer:

If carbon emissions stopped, the oceans catch up with the atmosphere, the Earth's temperature would rise about another 1.1F (0.6C).

Explanation:

8 0

3 years ago