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Lesechka [4]
1 year ago
8

A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.

Physics
2 answers:
jeka57 [31]1 year ago
7 0

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

Zarrin [17]1 year ago
4 0

Answer:

0

Explanation:

The normal force will be perpendicular to the motion. A zero-work situation is one where "a force everywhere perpendicular to the motion does no work" so according to this principle, no work is done.

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Which of the following statements describe the transfer of energy?
Lady_Fox [76]

The answer should be A weighlifter lifting 100lb. bell bar from his chest to a certain height.

7 0
2 years ago
The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel
aniked [119]

Answer:

= 2.52 x 10^ 6 m/s        

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv)  ÷ d

Here,  d = distance between the two plates

          q = charge of the charged particle

         Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a =  (q) (Δv)  ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv)  ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv)  ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 =  ut + 1/2 at^2 [proton}

put d/2 =  ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron)  ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

   Note, e = 1.60 x 10^-19

           m(electron) = 9.11 X 10^-31

            m(proton) = 1.67  X 10^-27

Input these values into the formula above, initial speed, UI =  

           = 2.52 x 10^ 6 m/s          

7 0
3 years ago
A particle moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.
mrs_skeptik [129]

The particle's acceleration is 5.1 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²

Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s

The given parameters are;

  • V1 = 2.35 m/s
  • V2 = - 8.72 m/s
  • T1 = 3.42s
  • T2 = 5.59s

Acceleration a = ΔV ÷ ΔT

a = (2.35 + 8.72) / (5.59 - 3.42)

a = 11.07 / 2.17

a = 5.1 m/s²

Therefore, the particle's acceleration is 5.1 m/s²

Learn more about Acceleration here: brainly.com/question/9069726

#SPJ1

4 0
1 year ago
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
olya-2409 [2.1K]

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

3 0
2 years ago
If mass is the amount of MATTER present in the
Korvikt [17]

Answer:

Yes it would be different on Earth and the moon

6 0
2 years ago
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