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umka2103 [35]
3 years ago
15

A running back holds 0.25 kg football and runs 100m down the football field. In addition to scoring six points for his team, he

also does ___ work on the football
A) no
B) 25 J of
C) 100 J of
D) 250 J of

Select the option which will cause a worker to do the maximum amount of work.
A) Using a lever to lift 100 newtons up to 4 meters on to a shelf
B) Using her hands to lift 200 newtons up 1 meter on to a table
C)Using a ramp to lift a 200 newton box up 1 meter into a truck
D) Using a pulley to lift to 100 newton box up 3 meters onto a platform

The power resulting from 420 Nm of work being completes in 3 seconds
A) 140 Joules
B)140 Newton-meters

What do you know about the law of conservation of energy? Check all the statements that are true

A)The total amount of mass in a system remains constant regardless of the changes that take place in system
B) Energy is neither creates or destroyed, it only changes form
C) Half-way down the ramp, PE = 50J and KE = 50J
D) In a closed system, a system that isolated from its surrounds, the total energy of the system is conserved
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

<h2>1) there is no work done on the system</h2><h2>2) A) Using a lever to lift 100 newtons up to 4 meters on to a shelf</h2><h2 /><h2>3) P = 140 W</h2><h2>4) D) In a closed system, a system that isolated from its surrounds, the total energy of the system is conserved</h2>

Explanation:

1) As we know that work done is the product of force and the displacement of the point of action where force is applied

So here we have

W = 0

as there is no displacement in the direction where the force is applied

2)As we know that work is product of force and displacement

So we will have

W_1 = 100 \times 4 = 400 J

W_2 = 200 \times 1 = 200 J

W_1 = 200 \times 1 = 200 J

W_1 = 100 \times 3 = 300 J

So maximum work is done on

A) Using a lever to lift 100 newtons up to 4 meters on to a shelf

3)

As we know that power is rate of work done

so we have

P = \frac{W}{t}

P = \frac{420}{3}

P = 140 W

4)

As per energy conservation we know that

D) In a closed system, a system that isolated from its surrounds, the total energy of the system is conserved

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a. The particle has position vector

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\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath

b. Its velocity vector is equal to the derivative of its position vector:

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c. At t=7.00\,\mathrm s, the particle has position

\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath

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Then its speed at this time is

\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}

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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
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Where,

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Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

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3 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: <em>The total amount of energy transfer by the work done is 54.81 kJ.</em>

Part b: <em>The total amount of energy transfer by the heat is 54.81 kJ</em>

Explanation:

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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

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Part b:

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

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