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frutty [35]
3 years ago
5

A box with a mass of 12.5 kg sits on the floor. how high would you need to lift it for it to have a gpe of 355 j

Physics
1 answer:
grandymaker [24]3 years ago
3 0
Gpe=mgh

355=12.5×10×h
h=355/125= 2.84m
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Answer:

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<em>weight</em><em>=</em><em>7</em><em>5</em><em>×</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>3</em><em>5</em><em>N</em>

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4 0
3 years ago
Example of the center of the gravity<br>​
velikii [3]

Answer:

The example of the center of the gravity is the middle of a seesaw

Explanation:

I hope this will help you and plz mark me brainlist

5 0
3 years ago
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Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

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Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

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Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

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3 × 19.6 = u²(0.7071)²

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Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

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T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
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damaskus [11]

Answer:

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Explanation:

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3 0
3 years ago
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vekshin1
They will be travelling slower than 10mph.
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maybe they are travelling at 5mph but I'd say it's a safer option to chose under 10mph
4 0
3 years ago
Read 2 more answers
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