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a_sh-v [17]
3 years ago
6

A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the waterme

lon from the roof to the ground.
Physics
1 answer:
Nonamiya [84]3 years ago
7 0

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

W=4.8\ kg\times 9.8\ m/s^2\times 18\ m

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

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A bus is traveling with a uniform acceleration of 2.75 meters/second2. If the initial velocity of the bus is 16.5 meters/second,
Kaylis [27]
<span>The velocity will be 41.25 m/s2 after 9 seconds. To find velocity after a specific time period, multiply the acceleration (2.75) times the number of seconds (9) to receive 24.75 m/s, then add that to the initial velocity of 16.5 m/s. 24.75 + 16.5 = 41.25 m/s2.</span>
8 0
3 years ago
An 85.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 28.0°
irina [24]

Answer:

A  75.1 N and a direction of 152° to the vertical.

B 85.0 N at 0° to the vertical.

Explanation:

A) The interaction partner of this normal force has what magnitude and direction?

The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>

B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.

Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.

<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>

8 0
2 years ago
A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He
nikklg [1K]

Answer:

0.9432 m/s

Explanation:

We are given;

Mass of swimmer;m_s = 64.38 kg

Mass of log; m_l = 237 kg

Velocity of swimmer; v_s = 3.472 m/s

Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.

So;

Initial momentum = final momentum

m_l × v_l = m_s × v_s

Where v_l is speed of the log relative to water

Making v_l the subject, we have;

v_l = (m_s × v_s)/m_l

Plugging in the relevant values, we have;

v_l = (64.38 × 3.472)/237

v_l = 0.9432 m/s

8 0
3 years ago
Suppose the electrostatic force between two electrons is F. What is the electrostatic force between an electron and a proton
mars1129 [50]

Answer:

D. −F

Explanation:

the rest of the answers are

2/3F

The force is represented as a positive quantity and is repulsive.

Electrostatic force is inversely proportional to the square of the distance.

The direction of the force changes, and the magnitude of the force quadruples.

hope this helps sorry if i was too late! :)

8 0
3 years ago
A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.1
Stolb23 [73]

Answer:

it could be 69

Explanation:

because it is a solution

4 0
3 years ago
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