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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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Hell please thanks!!!!!!’
Slav-nsk [51]

Answer:

liquid phase

Explanation:

it is liquid phase because molecules are not that tightly packed as solid and not that far away from each other as in gas phase.

5 0
3 years ago
1. Why do only some people get addicted to<br> drugs?
nasty-shy [4]

Answer: All drugs of abuse trigger a surge of dopamine

Explanation: It creates a rush of “wanting” in the brain. This makes us crave more drugs.

4 0
3 years ago
Read 2 more answers
D. What is the net force on the bowling ball rolling lane
3241004551 [841]

Answer:

Friction

Explanation:

3 0
3 years ago
A thin ring of radius R is uniformly charged with the total positive charge Q. 1. Find the electric field at distance h right ab
Paha777 [63]

Answer:

E_{net} = \frac{k Q h}{(R^2 + h^2)^{\frac{3}{2}}}

Explanation:

If charge Q is uniformly distributed over the ring then for small element of the ring the charge will be given as

dq = \frac{Q}{2\pi R} dL

now electric field due to this small element on the axis of ring at height "h" is given as

dE = \frac{k dq}{R^2 + h^2}

now since this electric field is radially in the direction of joining the position of charge to the given point

so net field due to ring will be along the axis of ring

so we have

E_{net} = \int dEcos\theta

E_{net} = \int \frac{k dq}{R^2 + h^2} \frac{h}{\sqrt{R^2 + h^2}}

E_{net} = \frac{k Q h}{(R^2 + h^2)^{\frac{3}{2}}}

4 0
3 years ago
An object that is initially at rest on a frictionless floor is struck by a flying rock with an impulse jfonox. what is the final
jasenka [17]

Answer:

jfonox

Explanation:

since impulse is the change in momentum given by

Impulse = Force x time = change in momentum

I = ft = mv-mu= m(v-u)= jfonox

since the object is initially at rest

u=0,

mv is the final momentum of the object

I =Ft = m(v - 0) = jfonox

mv= jfonox

this show that the final momentum of the object at rest is equal to the impulse received

if  any of the mass m, final velocity v or the impact force and the duration of impact of the object is known then we can find or quantify the final momentum.

5 0
3 years ago
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