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Aleks04 [339]
2 years ago
14

9. a. Determine the mass of a football which has a weight of 0.80 N on a planet where the gravitational field strength is 2 N/kg

. (3 marks)
​
Physics
1 answer:
krek1111 [17]2 years ago
5 0

Answer:

m = 0.4 [kg]

Explanation:

Weight is considered as a force and this is equal to the product of mass by gravitational acceleration.

W=m*g\\

where:

W = weight = 0.8 [N]

m = mass [kg]

g = gravity acceleration 2[N/kg]

Therefore:

m=W/g\\m = .8/2\\m = 0.4 [kg]

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D. Nucleus because it is not a part of the group.
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The Earth’s internal __________ source provides the energy for our dynamic planet, providing it with the driving force for on-go
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Students have been assigned to write reports on cell organelles. Eric’s report is about the organelle that supports and gives st
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3 years ago
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From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o
maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

tangent( O ) = \frac{opposite}{adjacent}

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

tangent( 4) = \frac{53 feets}{DB}

DB=  \frac{53 feets}{tangent( 4)}

DB=  757,93 feets

This its equal to the distance from the boat to the shoreline.

4 0
2 years ago
1000 kg of water initially at 6 m/s runs through a hydro-generator. If the water leaves the generator at velocity of 4 m/s, and
r-ruslan [8.4K]

Answer:

8.00 kJ

Explanation:

The first thing is to determine what quantities are there.

the mass of water = 1 000 kg

initial velocity, u = 6 m/s

final velocity, v = 4 m/s

the generator is operating at 100 % efficiency, so there is no energy loss.

The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.

The kinetic energy is calculated as follows:

Ek = 1/2 mv²

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    = 8 000 J/s

    = 8.00 kJ  Ans

4 0
3 years ago
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