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____ [38]
2 years ago
7

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

al force required to keep the faster car on the road without skidding is F, then the frictional force required to keep the slower car on the road without skidding is __-
Physics
1 answer:
yawa3891 [41]2 years ago
8 0

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0
2 years ago
in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at
Tema [17]
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

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3 0
2 years ago
¿Alguien me puede ayudar? Problema: Un niño le pide gastada a su papá y éste le contesta que le dará los $120 que tiene en su bo
VLD [36.1K]

Answer: there are 15 coins of $2 and 18 coins of $5

Explanation:

I will answer in English.

X is the number of $5 coins.

Y is the number of $2 coins.

We have the system of equations:

Y + X = 33

Y*2 + X*5 = 120

first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:

Y = 33 - X.

Then we can replace it in the other equation:

(33 - X)*2 + X*5 = 120

66 - X*2 + X*5 = 120

X*3 = 54

X = 54/3 = 18

and using the equation for Y.

Y = 33 - X = 33 - 18 = 15

So there are 15 coins of $2 and 18 coins of $5

3 0
2 years ago
9ma electric current is flowing through a conducting wire , then the number of electron passing through it in 3 min is
Leviafan [203]

Answer:

1.0125 x 10^19

Explanation:

current flowing through conductive wire= 9mA = 9 x 10^ -3 A

charge passing per 3 min

Q = It

= 9 x 10^ -3 x (3 x 60)

= 1.620 C

no of electrons in charge

Q = ne

1.620 = n x 1.6 x 10 ^ -19

n. = 1.0125 x 10 ^19

4 0
2 years ago
A circular coil has a 10.0 cm radius and consists of 30.0 closely wound turns of wire. an externally produced magnetic field of
IrinaVladis [17]
Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:

ΦB = BA
          A = πr^2
          A = π(.1 m)^2
          A = π/100 m^2   

 ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)

The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:

L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH

6 0
2 years ago
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