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Anvisha [2.4K]
3 years ago
13

You are at the edge of a diving board that is 9 meters above the water. If you weigh 500 Newtons, what is your potential energy?

Physics
2 answers:
Semenov [28]3 years ago
6 0

Answer:

4500 J

Explanation:

First, let's define some equations and derivations.

Our potential energy formula is:

  • \displaystyle U = mgh

Where <em>m </em>is mass (in kg), <em>g</em> is the gravitational constant (in m/s²), and <em>h</em> is height (in m).

We also know that <em>mg</em> is equal to the weight of an object (in N), from Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration).

Therefore, we can simply substitute force into the equation:

  • \displaystyle U = Fh

Where <em>F</em> is the force (in N) and <em>h</em> is still height (in m).

Now we can calculate the amount of potential energy in our system, measured in joules.

Substitute in the given variables, F = 500 N and h = 9 m:

  • \displaystyle U = (500 \ N)(9 \ m)

Using simple Pre-Algebra rules, we find that:

  • \displaystyle U = 4500 \ J

This tells us that the we have 4500 joules of potential energy when I am 9 meters above the water on the edge of the diving board.

ICE Princess25 [194]3 years ago
6 0

Answer:

4500 J

Explanation:

First, let's define some equations and derivations.

Our potential energy formula is:

Where m is mass (in kg), g is the gravitational constant (in m/s²), and h is height (in m).

We also know that mg is equal to the weight of an object (in N), from Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration).

Therefore, we can simply substitute force into the equation:

Where F is the force (in N) and h is still height (in m).

Now we can calculate the amount of potential energy in our system, measured in joules.

Substitute in the given variables, F = 500 N and h = 9 m:

Using simple Pre-Algebra rules, we find that:

This tells us that the we have 4500 joules of potential energy when I am 9 meters above the water on the edge of the diving board.

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Answer:

Push - The most common form of force is a push through physical contact (like a lawnmower or shopping cart)

Pull - You can apply a force by directly pulling on an object (like pulling a wagon)

Explanation:

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A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t
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The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

p_i = (M+m)u

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

p_f = MV+mv

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

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4 years ago
Does a feather fall as fast as a rock in a vacuum? If so why?
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Answer:

No.

Explanation:

A feather is less dense and thus less force exerted while a rock is very dense thus exerting more force .

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You plug in an extension cord and you have to be very careful around electrical outlet. However you can handle his extension cor
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A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0
3 years ago
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