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3 years ago

Which of these will be the correct relationship between work input and work output?

Physics

1 answer:

dsp733 years ago

8 0

<u>Answer:</u>

Work input = Work output * Work against friction is your answer so C

<u>Explanation:</u>

I hope this helps you :)

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Which of the following are not correctly balanced reactions?

djyliett [7]

aaaaaaaaaaaaaa a or c

3 0

3 years ago

Read 2 more answers

PLEASE HELP

Mnenie [13.5K]

Velocity= 1/4 = 0.25 m/s towards the ball

7 0

3 years ago

A truck traveling at a constant speed of 24 m/s passes a more slowly moving car. The instant the truck passes the car, the car b

Ymorist [56]

Answer:

7.15 m/s

Explanation:

We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.

The truck moves at constant speed, we can use the equation for position under constant speed:

Xt = X0 + v*t

The car is accelerating with constant acceleration, we can use this equation

Xc = X0 + V0*t + 1/2*a*t^2

We know that both vehicles will meet again at x = 578

Replacing this in the equation of the truck:

578 = 24 * t

We get the time when the car passes the truck

t = 578 / 24 = 24.08 s

Before replacing the values on the car equation, we rearrange it:

Xc = X0 + V0*t + 1/2*a*t^2

V0*t = Xc - 1/2*a*t^2

V0 = (Xc - 1/2*a*t^2)/t

Now we replace

V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s

6 0

3 years ago

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

3 years ago

A clindrical rod of uniform density is located with its center at the origin, and its axis along the axis. It rotates about its

zhenek [66]

Answer:

6093.2328 J

Explanation:

For cylindrical rod moment of inertia will be

$I_X=I_Y=\frac{1}{12}m(3r^2+h^2)$

$I_X=I_Y=\frac{1}{12}\times 4(3\times 0.06^2+0.6^2)=0.1236$ $kg -m^2$

we have given time =0.02 sec

Angular speed = $\frac{2\pi }{T}=\frac{2\times 3.14}{0.02}=314\ rad/sec$

Rotational KE = $\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.1236\times 314^2=6093.2328\ J$

5 0

3 years ago