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grin007 [14]
3 years ago
11

A 1900kg airplane is flying at an altitude of 510 above the ground. What is the gravitational potential energy in Joules?

Physics
1 answer:
Natalija [7]3 years ago
8 0

Answer: 9496200 joules

Explanation:

Gravitational potential energy, GPE is the energy possessed by the moving plane since it moves against gravity.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

Since Mass = 1900kg

g = 9.8m/s^2

h = 510 metres (units of height is metres)

Thus, GPE = 1900kg x 9.8m/s^2 x 510m

GPE = 9496200 joules

Thus, the gravitational potential energy of the airplane is 9496200 joules

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Answer:

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2 years ago
An object moves with a constant velocity of 17 m/s for 200 s. How far did it move during that time?
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3 years ago
It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you
Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used  to solve this question.

From the diagram, the horizontal component of the force is:

F_X = F_{cos \ \theta}

Replacing 42°  for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

On the other hand, the vertical component  is ;

F_Y = Fsin \ \theta

Replacing 42°  for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21  \ N

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

f = \mu \ N

Where;

\mu is said to be the coefficient of the friction

N = the  normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing F_Y \ for \ F_{sin \  \theta}. The normal reaction can now be:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

F_X = \mu \ ( mg - \ F_Y)

Making \mu the subject of the formular in the above equation; we have the following:

\mu \ = \ \frac{F_X}{mg - F_Y}

Replacing the following values: i.e

F_X \ = \ 64.65 \  N

m = 73 Kh

g  = 9.8 m/s²

F_Y = \ 58.21 N

Then:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Thus, the coefficient of friction is = 0.0984

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The diagram is an illustration of<br> NO LINKS
V125BC [204]

Answer:

A longitudinal wave

Explanation:

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