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3 years ago

What are some examples of static potential energy?

Physics

1 answer:

creativ13 [48]3 years ago

6 0

Answer:

rubbing your socks on carpet

rubbing your hair with glove

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a phenomenon whereby the moon appears bluish owing to smoke or dust particles in the atmosphere

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3 years ago

What is the relationship between environmental science and public policymakers?

jekas [21]

In a simple sentence I would say using environmental science public policy maker would be able to create policies on environmental issues such as littering,recycling specific items/reusing,not wasting excess energy,ect.

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3 years ago

Air, water and salt solution are given in the table.

Oxana [17]

Answer:

I need help

Explanation:

Albert tries to move a 200 N car and it moves 0 m. A tow truck tries to move the 200 N car and it moves 2 m off the ground onto the truck bed. Who put more work into moving the car? Explain your response using the space below. Be sure to show your calculations. *

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3 years ago

A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the

MaRussiya [10]

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

$v^{2}=u^{2}+2as$ and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

$u=\sqrt {v^{2}-2gs}$

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

$u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s$

5 0

3 years ago

An experimental rocket designed to land upright falls freely from a height of 2.59 102 m, starting at rest. At a height of 86.9

aleksandr82 [10.1K]

Answer:

The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²

The rocket's motion for analysis sake is divided into two phases.

Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m

Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.

Explanation:

The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.

The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.

The detailed step by step solution to the problems can be found in the attachment below.

Thank you and I hope this solution is helpful to you. Good luck.

5 0

3 years ago