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skelet666 [1.2K]
3 years ago
12

A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a

60.0 N force for the first 7.09 m, at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 7.09 m. How much total work did the person do on the box?

Physics
1 answer:
gogolik [260]3 years ago
3 0

Answer:

W= 638.1 J

Explanation:

As we know that

Work done is the area of the force and displacement diagram.

W=∫F.dx

W=Work

F=force

dx=Elemental displacement

From the diagram ,area A

A= 60 x 7.09 + 1/2 x 60 x 7.09

A= 638.1 J

So the work W

W= 638.1 J

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If a sound travels 343m/s through air and has a frequency of 800Hz, what would the wavelength be of the same sound traveling thr
VARVARA [1.3K]

Answer:

0.625m

Explanation:

Now Velocity of a wave ,V = frequency × wavelength

Wavelength =velocity /frequency

=500/800 =0.625m

7 0
2 years ago
Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f
Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

\lambda=380 nm =3.8\cdot 10^{-7}m (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

Substituting,

E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J

And keeping in mind that

1 eV = 1.6\cdot 10^{-19}J

This energy converted into electronvolts is

E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0
3 years ago
a stone is dropped from the top of 50 m high tower simultaneously another stone is thrown upward with a speed of 20 m/s . calcul
Darina [25.2K]
'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.

For the stone dropped from the 50m tower:

H = +50 - (1/2) G T²

For the stone tossed upward from the ground:

H = +20T - (1/2) G T²

When the stones' paths cross, their <em>H</em>eights are equal.

50 - (1/2) G T² = 20T - (1/2) G T²

Wow !  Look at that !  Add (1/2) G T² to each side of that equation,
and all we have left is:

50 = 20T  Isn't that incredible ? ! ?

Divide each side by 20 :

<u>2.5 = T</u>

The stones meet in the air 2.5 seconds after the drop/toss.

I want to see something: 
What is their height, and what is the tossed stone doing, when they meet ?

Their height is  +50 - (1/2) G T² = 19.375 meters

The speed of the tossed stone is  +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it.  The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.
7 0
3 years ago
The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will __
Kipish [7]

Answer:

The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will decrease with time, the surface area between the tire and the road will increase with time, and the net force the tire exerts on the road will be constant with time.

Explanation:

when a wheel has an air leak, it means that the inside of the tire has less air, which means that there will be less air pushing the walls of the tire so that the air pressure decreases.

On the other hand, the tire begins to deform due to lack of air which increases the area of ​​contact with the floor.

As the weight of the car remains constant and the air has a negligible mass the force towards the road will be the same

4 0
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Genrish500 [490]

Answer:

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Explanation:

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6 0
3 years ago
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