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deff fn [24]
3 years ago
11

A 67kg stove is located on the 70th floor of a building (230m above ground). The stove has _______ energy. Calculate it.

Physics
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

1.51\cdot 10^5 J of gravitational potential energy

Explanation:

the gravitational potential energy of an object is given by:

U=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the heigth of the object above the ground

In this problem, we have:

- m = 67 kg is the mass of the stove

- g = 9.8 m/s^2 is the acceleration due to gravity

- h = 230 m is the height of the stove above the ground

Substituting into the equation, we find

U=(67 kg)(9.8 m/s^2)(230 m)=1.51\cdot 10^5 J

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Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\sqrt{\dfrac{k}{m}}

Where, m = mass

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Put the value into the formula

\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}

\omega=4.74\ rad/s

We need to calculate the period of oscillation,

Using formula of time period

T=\dfrac{2\pi}{\omega}

Put the value into the formula

T=\dfrac{2\pi}{4.74}

T=1.33\ sec

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4 0
3 years ago
In the graph below, why does the graph stop increasing after 30 seconds?
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Answer:

The answer is "Option C".

Explanation:

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6 0
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Which planet has the GREATEST attraction to the sun?
AveGali [126]
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Use your data (attachment) from Part 3 and Newton’s laws to explain why the force meter measures a force if the cart is moving a
vazorg [7]
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3 years ago
Calculate the energy used by a radio of power 30W in 1 minute.
Alekssandra [29.7K]

Answer:

<u>1.8kJ</u>

Explanation:

Formula :

<u>Energy used = Power x time</u>

<u />

===============================================================

Given :

⇒ Power = 30 W

⇒ Time = 1 minute = 60 seconds

=============================================================

Solving :

⇒ Energy used = 30 W × 60 s

⇒ Energy used = 1,800 J

⇒ Energy used = <u>1.8kJ</u>

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2 years ago
Read 2 more answers
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