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2 years ago

Why would a skier try to lower his center of gravity?

1 answer:

8 0

He would lower his center of gravity so that it would be harder for him to fall, and so that he can stay and become balanced on his ski's.

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Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

Anton [14]

Answer:

Explanation:

moment of inertia of each blade which is similar to rod rotating about its one end

= 1/3 ml²

moment of inertia of 3 blades = ml²

= 5500 x 46²

I = 11638 x 10³ kg m²

angular velocity = 2πn where n is rotation per second

n = 11 / 60

angular velocity = 2π x 11/60

= 1.1513 rad /s

angular momentum

= moment of inertia x angular velocity

= 11638 x 10³ x 1.1513

= 13399 x 10³ kg m² per second.

4 0

3 years ago

A child pulls a toy wagon with a force of 25.0 N for a distance of 8.5M. How much work did the child do on the wagon?

zzz [600]

Answer:

f=25.0 w=d 25.0÷8.5

s=8.5m

7 0

2 years ago

A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a friction

marissa [1.9K]

Answer:

Part a: The total amount of energy transfer by the work done is 54.81 kJ.

Part b: The total amount of energy transfer by the heat is 54.81 kJ

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K$

Volume of the tank is given as

$V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3$

Final mass is given as

$m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg$

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\\m=3-2=1 kg$

The initial mass in the cylinder is given as

$m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg$

The mass after the process is

$m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\$

Now the volume 2 of the cylinder is given as

$V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3$

Part a:

So the Work done is given as

$W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ$

The total amount of energy transfer by the work done is 54.81 kJ.

Part b:

The total energy transfer by heat is given as

$Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ$

As the temperature is constant thus change in internal energy is 0.

The total amount of energy transfer by the heat is 54.81 kJ

7 0

2 years ago

At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assum

Verdich [7]

Answer:

Ec = 6220.56 kcal

Explanation:

In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.

You use the following formula:

$W_c=Mgh$ (1)

Wc: work done by the climber

g: gravitational constant = 9.8 m/s^2

M: mass of the climber = 78.4 kg

h: height reached by the climber = 5.42km = 5420 m

You replace in the equation (1):

$W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J$ (2)

Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:

$0.16(E_c)=4,164,294.4J$