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Diano4ka-milaya [45]
2 years ago
6

How does the position of an object relate to the energy stored in an object?

Physics
1 answer:
Kamila [148]2 years ago
4 0

Answer:

Potential Energy

Explanation:

Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

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How is acceleration calculated?
steposvetlana [31]

Answer:

Acceleration is the change in velocity divided by time

Explanation:

This is the correct answer because distance divided by time is the position. Speed multiplied by time is the distance. And acceleration is not just velocity, but the change in velocity over time.

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What is newton's 3rd law of physics ​
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Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

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A doctor has a patient blow on one side of a U-shaped tube that is partially
mart [117]

Answer:

A. 150 Pa

Explanation:

Answer is 150 Pa

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3 years ago
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
Freeeeeeeeeeeeeeeeeeeeeeeee points
andreyandreev [35.5K]

Answer:

yehyehhfujghnnkoohu

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2 years ago
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