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Musya8 [376]
3 years ago
5

Suppose a 4000 kg elephant is hoisted 20m above Earth's surface. What will be the elephant's gravitational potential energy?

Physics
1 answer:
inna [77]3 years ago
8 0
We know, U = m * g * h
U = 4000 * 9.8 * 20
U = 784000 J

In short, Your Answer would be: 784000 J or 784 KJ

Hope this helps!
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Argon, Lithium and Potassium

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Emerging adults are typically financially independent from their parents.<br> True or false?
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False

Explanation:

According to USA today "Though nearly 60% of the 18 to 21 year olds surveyed said they have a job, and 70% of the 22 to 26 year olds said the same, many seem to rely on mom and dad to fund non-essential expenses. Nearly 75% of the 18 to 21 year olds and more than 40% of 22 to 26 year olds said their parents pay their monthly cell phone bill. Just 54% of 22 to 26 year olds own or lease their own car; and 43% of 22 to 26 year olds are living at home with their parents."

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A kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward what is the distance and magnitu
Elina [12.6K]

Answer:

Distance = 64 metres

Displacement = 28 metres

Direction = northward

Explanation:

Given that a  kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward

Distance covered is only about the magnitude of the length covered.

Distance = 22 + 18 + 24 = 64 metres

The direction will be considered when calculating the displacement

Let northward be positive and southward be negative

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Since the displacement is positive, the direction of the motion is northward.

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3 years ago
What is the relationship between mass and weight?
Vikentia [17]

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C. Weight is mass times the acceleration of gravity.

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4 0
3 years ago
A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its
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Answer:

a). E_{kA}=1.2635 J

b). V_{B}=4.535\frac{m}{s}

c). ΔE_{t}=8.4635 J

Explanation:

ΔE=kinetic energy

a).

E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J

b).

E_{kB}=\frac{1}{2}*m*v_{B} ^{2}

V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}

c).

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E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J

3 0
3 years ago
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