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olga nikolaevna [1]
3 years ago
14

a 2.80 kg mass is dropped from a height of 4.50 m. find its potential energy(PE) at the moment it is dropped. PLEASE HELP

Physics
1 answer:
Maurinko [17]3 years ago
4 0

Answer:

126 J

Explanation:

The potential energy only depends on the vertical height from the ground level.

We consider the ground level to have zero P.E.

We get, when it is 4.5 m above the ground level,

P.E. =  mgh

      = 2.8×10×4.5

      = 126 J

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Answer:

a) -35.6°C

b) 237.4 K

Explanation:

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T_c=\frac{5}{9}(T_f-32)

a) Therefore to convert -32°F to celsius, substitute it into the celsius

\begin{gathered} T_c=\frac{5}{9}(-32-32) \\  \\ T_c=\frac{5}{9}(-64) \\  \\ T_c=-35.6^0C \end{gathered}

b) To covert to the Kelvin scale, use the formula below

\begin{gathered} T_k=T_c+273 \\  \\ T_k=-35.6+273 \\  \\ T_k=237.4K \\  \end{gathered}

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1 year ago
An elevator car has a mass of 750 kg, and its three passengers have a combined mass of 135 kg. If the elevator and its passenger
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Answer:

The change in gravitational potential energy is -1.80x10⁵ J.

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The change in gravitational potential energy is given by:

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Where:

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h: is the height

For the car and the passengers we have:

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The minus sign is because when the elevator car and the passengers are up they have a bigger gravitational potential energy than when they are in the ground.

Therefore, the change in gravitational potential energy is -1.80x10⁵ J.

I hope it helps you!                                                

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3 years ago
A skillet is removed from an oven whose temperature is 450°F and placed in a room whose temperature is 70°F. After 5 minutes, th
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Answer:

attached below

Explanation:

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3 years ago
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Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

Distance between charges , d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.

Magnitude of charges , q=1\ nC = 10^{-9}\ C.

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Putting all values and r=12\times 10^{-2}\ m.

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Case B) (x,y) = (0 cm, 12.0 cm) :

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Answer:

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So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0
3 years ago
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