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Liula [17]
2 years ago
13

you push a block of 50kg with a force of 400 N over a distance of 8m. How much kinetic energy does the box gain?

Physics
1 answer:
MatroZZZ [7]2 years ago
4 0

Answer:

160000

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
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Explanation:

Orbital speed= 2pi x radius / time period

=2pi x 1.5x10^11 / 365.25

=2.58x10^9m/day

8 0
3 years ago
0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for
Paladinen [302]

Assuming the accleration applied was constant, we have

v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)

\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)

\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}

Then the force applied to the ball is given by

F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)

\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N

8 0
2 years ago
I need help with this. I think I got it correct, but I wanna make sure.
kicyunya [14]

Answer:

A because the bigger it is the the more force needs to act apond it

Explanation:

3 0
3 years ago
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Machines A, B, and C do the same amount of work, but they each take a different amount of time. Machine A takes more time as com
Amiraneli [1.4K]
So power is considered as the rate of doing work. Base on the problem given, my analysis is that the machine who finish the work faster is machine C. Therefore, in order to finish the same amount of work in a short period of time you are going to expend the most power. My answer is Machine C. 
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3 years ago
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A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m × 0.55 m. The magnetic field has
Pani-rosa [81]

Answer:

Part a)

EMF = 0.38 V

Part b)

\frac{dA}{dt} = 0.43 m^2/s

Explanation:

Part a)

Initial value of magnetic flux is given as

\phi_1 = BAcos\theta

\phi_1 = (2.1)(0.35 \times 0.55) cos65

so we have

\phi_1 = 0.17 Wb

Final flux through the loop is given as

\phi_2 = 0

now EMF is given as

EMF = \frac{\phi_1 - \phi_2}{\Delta t}

EMF = \frac{0.17 - 0}{0.45}

EMF = 0.38 V

Part b)

If magnetic field is constant while Area is changing

So EMF is given as

E = Bcos65 \times \frac{dA}{dt}

0.38 = 2.1 cos65(\frac{dA}{dt})

\frac{dA}{dt} = 0.43 m^2/s

5 0
2 years ago
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