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Scrat [10]
3 years ago
10

A car traveling on a level road initially has 440 kJ of mechanical energy. After the brakes are applied for a few seconds, the c

ar has only 110 kJ of mechanical energy. What best accounts for the missing mechanical energy
Physics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer:

Explanation:

Given that

There's a change in friction from 440 kJ to 110 kJ

This change is as a result of conversion of energy. The energy of the car system converted the mechanical energy in it, into a corresponding heat energy. During the course of this conversion, is where a whopping 330 kJ of energy went missing.

The 330 kJ of energy was lost as a result of the conversion of mechanical energy into heat energy by the use of friction.

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Two like-charged particles are placed close to each other. How would the force of repulsion be affected if the charge on one of
astra-53 [7]

Answer: It will remain the same

Explanation:

According to <u>Coulomb's Law:</u>    

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".  </em>

<em />

Mathematically this law is written as:  

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:  

F_{E}  is the electrostatic force  

K is the Coulomb's constant  

q_{1}=q_{2}=q are the electric charges , which in this case have the same positive charge

d is the separation distance between the charges

Rewritting we have:

F_{E}=K\frac{q^{2}}{d^{2}}    (2)

Now, if the first charge is doubled:

q_{1}=2q_

And the second is reduced to a half:

q_{2}=\frac{1}{2}q

We will have the following:

F_{E}=K\frac{(2q)(\frac{1}{2}q)}{d^{2}} (3)

F_{E}=K\frac{q^{2}}{d^{2}} (4)

As we can see equation (4) is equal to equation (2), this means the force of repulsion between both charges will remain the same

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3 years ago
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How many meters do you cover in a 10 km (10-K) race? 
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10,000 meters: 1,000 meters in every kilometer.
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Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction betw
jarptica [38.1K]

Answer:

a)  W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d)     the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

          W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

          W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

         θ = 90         cos 90 = 0

        W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

       θ = 180      cos 190 = -1

        W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

            m g sin θ  L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀=  + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

               The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

               ΔU = -U₀

         the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

              W_Total = W_weight - W_roce

as the body moves down

              W_Total> 0

Therefore the total work is positive

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Women generally have a lower centre of gravity than men, contributing to greater stability. Men generally have more muscle mass in their upper bodies,
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