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prohojiy [21]
2 years ago
6

How long does it take a microwave of power 0.2kW to sue 10000 J of energy

Physics
1 answer:
olasank [31]2 years ago
5 0

Answer:

50s .

Explanation:

\frak{\pink{Given}}\begin{cases}\textsf{ The power of microvave is 0.2kW .}\\\textsf{ Amount of energy is 10000 J .}\end{cases}

Here the power of the microwave is 0.2kW . And as we know that ,

  • Rate of doing work is called power .

So from the definition , we have ;

\sf\longrightarrow Power =\dfrac{Work}{time}

  • Here the work done is equal to the energy consumed by the microwave i.e. 10000 J .So we can write it as ,

\sf\longrightarrow Power =\dfrac{Energy}{time}

\sf\longrightarrow 0.2kW = \dfrac{10^4 J }{t} \\

\sf\longrightarrow 0.2 * 1000 W =  \dfrac{10^4 J }{t}

Cross multiply ,

\sf\longrightarrow t = \dfrac{ 10^4 }{ 0.2 * 10^3}s=\dfrac{10^4}{2*10^2} s

Simplify ,

\sf\longrightarrow \boxed{\bf t = 50s}

<h3>Hence the time taken is 50s .</h3>
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Your roommate drops your wallet down to you from the third-floor window of your apartment, which is 11.5 m from the ground. What
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Answer:

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1.a bag is dropped from a hovering helicopter. the bag has fallen for 2 s. what is the ball's velocity at the instant its hittin
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1. The bag's velocity immediately before hitting the ground.

Recall this kinematics equation:

Vf = Vi + aΔt

Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)

a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)

Δt = 2s

Plug in the values and solve for Vf:

Vf = 0 + 9.81×2

Vf = 19.62m/s

2. The height of the helicopter.

Recall this other kinematics equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (bag is dropped starting from rest)

a = 9.81m/s² (acceleration due to gravity of the earth)

Δt = 2s

Plug in the values and solve for d:

d = 0×2 + 0.5×9.81×2²

d = 19.62m

3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s

Reuse the equation from question 2:

d = ViΔt + 0.5aΔt²

Given values:

d = 19.6m (height of the helicopter obtained from question 2)

Vi = 2m/s

a = 9.81m/s² (acceleration due to earth's gravity)

Plug in the values and solve for Δt:

19.6 = 2Δt + 0.5×9.81Δt²

4.91Δt² + 2Δt - 19.6 = 0

Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):

Δt = 1.8s, Δt = −2.2s

The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.

Δt = 1.8s

Now we can use this new value of Δt to get the velocity before hitting the ground:

Vf = Vi + aΔt

Given values:

Vi = 2m/s

a = 9.81m/s²

Δt = 1.8s (result from previous question)

Plug in the values and solve for Vf:

Vf = 2 + 9.81×1.8

Vf = 19.66m/s

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They are similar to the earth. Not relative to the outer planets

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