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ExtremeBDS [4]
2 years ago
9

What is the gravitational potential energy of a 0.1 kg apple sitting on top of a 12.5 m flagpole? Note: GPE= mgh. m= mass, g= ac

celeration due to gravity (9.8 m/s^2),
h= objects height above the ground.
Physics
1 answer:
Pachacha [2.7K]2 years ago
6 0
<span>GPE= mgh = 0.1 kg * 9.8 m/s^2  * 12.5m = 12.25 J</span>
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Two things you can do to increase the acceleration of an object
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You can decrease the mass, or you can increase the force applied to the object
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3 years ago
A sailboat travels a distance of 600 m in 40 seconds. What speed is it going?
Gekata [30.6K]

Answer:

15 miles /seconds

Explanation:

Distance = 600m

Time = 40 seconds

Speed=?

speed =  \frac{distance}{time}  \\ speed =  \frac{600}{40}

Simplify

\frac{600}{40}  =  \frac{60}{4}  \\  = 15

7 0
3 years ago
The practice of science can answer only scientific questions. And scientific questions guide the design of investigations. What
meriva

The scientific question can be supported by evidence that must be true of the possible answers to a scientific question. Option D is correct,

<h3>What is a Scientific hypothesis?</h3>

A scientific hypothesis must meet two criteria, A scientific hypothesis must be testable, and a scientific hypothesis must be falsifiable The basic idea of a hypothesis is that there is no predetermined outcome.

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They are accurate regarding the scientific question and can be substantiated by proof.

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Hence option D is correct.

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6 0
2 years ago
A team of engineering students is designing a catapult to launch a small ball at A so that it lands in the box. If it is known t
Murljashka [212]

Answer:

Explanation:

If the initial velocity is U

Then the horizontal component of the velocity is

Ux= Ucosθ

Then the range for a projectile is give as

R=Ux.t

Where t is the time of flight

The time of flight is given as

t=2USinθ/g

Therefore,

R=Ux.t

R=UCosθ.2USinθ/g

R=U^2×2SinθCosθ/g

Then, from trigonometric ratio

2SinθCosθ= Sin2θ

R=U^2Sin2θ/g

Given that θ=32° and g=9.81m/s^2

Then

R=U^2Sin2×32/9.81

R=U^2Sin64/9.81

R=0.0916U^2

Then, range is given by R=0.0916U^2

A=0.0916U^2.

T

The box is at a distance A from the point of projection. Then the range R=A

R=0.0916U^2

A=0.0916U^2

Then,

U^2=A/0.0916

U^2=10.915A

Then the initial velocity should be

U=√10.915A

U=3.3√A

8 0
3 years ago
A river is flowing south at a rate of 3 m/s. Steven can roe directly across the river if he aims the raft 30 degrees. What rate
levacccp [35]

Answer:

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

Explanation:

Given data,

The river flowing south at the rate, v = 3 m/s

To reach the other side directly across the river, he aims the raft, Ф = 30°

The speed of his raft across the river is given by the formula,

                                          V = v / Sin Ф

                                             = 3 / Sin 30°

                                              = 6 m/s

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

3 0
3 years ago
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