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aleksklad [387]
4 years ago
13

A book is pushed across the table by a student. The student stops pushing on the book and the book slows down and stops. The for

ce stopping the book is a force known as
A) inertia.
B) friction.
C) the applied force.
D) the action-reaction force.
Physics
2 answers:
V125BC [204]4 years ago
6 0
B. Friction. If the book was touching nothing, there would be no force to change its velocity given to it by the initial push. Because the book is sliding across the table, the molecules in the table are "slowing down" the book because they are exerting the force of friction on the book. What happens is that the charges in the book's molecules are attracted to the charges in the table's molecules. The attracting force is in the negative direction, causing the book to experience friction and therefore slow down.
VashaNatasha [74]4 years ago
3 0
A. inertia 

Because it has to do with the motion of something, especially if it changes its pace. In this example, the book's motion, when sliding on the table, decreased because of less force being given off from the student. 
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If you could help me please.
11Alexandr11 [23.1K]

1) Does a 1 kg object weight 9.8 newtons on the moon? why?

<em>No.</em> 1kg of mass does not weigh 9.8N on the moon.

Weight = (mass) x (gravity).

Gravity is 9.8 m/s² on Earth, but<em> gravity is only 1.62 m/s² on the moon</em>.

2) How much does a 3-kg object weigh (on earth) in newtons?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (3 kg) x (9.8 m/s² )

<em>Weight = 29.4 N</em>

3) How much does a 20-kg object weigh (on earth) in newton?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (20 kg) x (9.8 m/s² )

<em>Weight = 196 N</em>

4) What must happen for the mass of an object to change?

When an object moves, its mass increases.  The faster it moves, the greater its mass gets.  But this is all part of Einstein's "Relativity".  The object has to move at a significant fraction of the speed of light before any change can be noticed or measured.  So as far as we are concerned, in everyday life, <em>the mass of an object doesn't change</em>, no matter where it is, or what you do to it.

5) What are 2 ways the weight of an object can change?

First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.

But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood.  So the weight can change even though the mass doesn't.

The weight of an object changes if you take it to a place where gravity is stronger or weaker.

Let's say we have an object whose mass is 90.72 kilograms.  Like me !    

As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons  (200 pounds).

. . . Fly me to the moon. Gravity = 1.62 m/s²  Weight = 147 Newtons (33 lbs)

. . . Drag me to Jupiter.  Gravity = 24.8 m/s²  Weight = 2,249 N (506 pounds)

My mass never changed, but my weight sure did.

8 0
3 years ago
A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad
maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

F = Force/Weight

A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}

F = 38272.5N

Therefore the Weight of the object is 3.82kN

4 0
3 years ago
Suppose a 500 mb chart valid today at 12 Z indicates a large trough over the eastern US and a large ridge over the western US. A
Zinaida [17]

Answer:

An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will __LOSE___ altitude if the altimeter is not corrected

7 0
3 years ago
CAN SOMEBODY PLEASE HELP ME! i need help and i wanna pass
umka21 [38]

Answer:it would be C

Explanation:

8 0
4 years ago
A plane monochromatic electromagnetic wave with wavelength ? = 3 cm, propagates through a vacuum. Its magnetic field is describe
Zepler [3.9K]

Answer:The Poynting vector SS represents the flow of energy in an EM field. Specifically, if uu is the energy density of the field, the Poynting vector satisfies the continuity equation for it:

∂u∂t+∇⋅S=0

∂u∂t+∇⋅S=0

in vacuum. (This is Poynting's theorem.)

In your particular problem, EE and BB are perpendicular and their cross product is proportional to the product of their amplitudes. Thus

Sz=cμ0B2.

Sz=cμ0B2.

You then have to use your knowledge of BB to work out SS.

Explanation:

7 0
3 years ago
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