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ivann1987 [24]
3 years ago
7

Which of the following best represents potential energy being converted to kinetic energy? A. A man jogs and stops to drink an e

nergy drink. B. A drawn bow is released, causing an arrow to fly across a field. C. A roller coaster rounds a curve and climbs the next hill. D. A tree is struck by lightning, and then it is set on fire.
Physics
2 answers:
Schach [20]3 years ago
3 0
<span>So we want to know which of the following is the best representation of converting potential energy into kinetic energy. The correct answer is C. A roller coaster rounds a curve to climb the next hill. So before he climbed the hill, the roller coaster had kinetic energy which he used to climb to the hill. Then the potential energy he has on the hill can again be transformed into kinetic energy when he will go down hill. </span>
igomit [66]3 years ago
3 0

Answer;

B.  A drawn bow is released, causing an arrow to fly across the field.

Explanation;

Potential energy is the energy in a body due to its position. While kinetic energy is the energy in a body due to its motion. The formula for potential energy is mgh, where m stands for mass, g stands for gravitational acceleration and h stands for height.

Potential energy is the stored energy that can become kinetic energy. Kinetic energy is given by 1/2mv², where m is the mass and v is the velocity.

Potential energy = Stored energy, usually energy in bonds, energy in height.  

Kinetic energy = Energy of movement

As you draw the bow, the elasticity of the bow stores the energy (potential), and once you release, the arrow flies across the field (kinetic).

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Why is the sky blue?
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3 years ago
Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o
KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹  N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹  N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea =  +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹  N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb =  -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴  +  -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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telo118 [61]

Answer

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Step-by-step explanation:

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An airplane has a speed of 135 mi/h in still air. It is flying straight north so that it is always directly above a north-south
vlabodo [156]

Answer:

The answer is below

Explanation:

Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.

The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph

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vₐ = vₐₙ

Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.

The direction of the wind θ is:

sin(θ / 2) = vₙ / 2vₐ

sin(θ / 2) = 70/ (2*135)

sin(θ / 2) = 0.2593

θ / 2 = sin⁻¹(0.2593) = 15

θ = 30⁰

2α = 180° - 30°

2α = 150°

α = 75°

a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.

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