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Korvikt [17]
3 years ago
10

Which object will have the most potential energy?

Physics
1 answer:
Svetradugi [14.3K]3 years ago
5 0

Answer:  the most potential energy ==  5 kg book, 2 m from the ground= 98 Joules

Explanation:

potential energy = m g h

m = mass

g = acceleration due gravity  = 9.8 m/s²

h = distance above  ground

1.  Pe₁ = 1 kg x 2 m x g  = 2 g

2. Pe₂ = 5 kg x 2 m x g = 10 g = 10 kg m x 9,8 m/s² = 98 Joules

3. Pe₃ = 1 kg x 0,5 m x g = 0,5 g

4. Pe₄ = 5 kg x 0.5 m x g = 2,5 g  

10 > 2,5 > 2 >0,5

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A 783kg elevator rises straight up 164 meters. What is the potential energy of the elevator?
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Answer:

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Explain why pulleys are in the lever family.
Aleks04 [339]

Answer:

The leverage or mechanical advantage of pulleys is less obvious, but you can "gang" multiple pulleys together into two sets (blocks) and run the ropes back and forth between the two sets to increase the number of lengths of rope running between them. One end of the rope is connected (fixed) to one of the blocks, and you get to pull on the other end after it is passed back and forth between the blocks of pulleys. This is sometimes called a block and tackle arrangement. With a hook on each side of the block set, you can move a heavy load much like levers do, by multiplying the force. You have to pull more rope just like you have to move a lever more on one side of the fulcrum as compared to the other. When you get all the rope pulled out that you can, you can not move the load anymore because you have become "two-blocked" which means the two blocks are together. Credits to: Moin Khan

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3 years ago
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

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 jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e 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 fina 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 ony 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
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Describe how the number of photoelectrons emitted from a metal plate in the photoelectric effect would change if the following o
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Answer:

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b)   λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,

c)  threshold energy

        h f =Ф

Explanation:

It's photoelectric effect was fully explained by Einstein by the expression

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Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b) wavelength is related to frequency

         λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.

c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy

        h f =Ф

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