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slega [8]
3 years ago
15

What is the maximum speed at which a car can safely travel around a circular track of radius 55.0 m If the coefficient of fricti

on between the tire and
road is 0.3502
2.60 m/s
4.39 m/s
13.7 m/s
43.0 m/s
Physics
1 answer:
Stella [2.4K]3 years ago
5 0

Answer:

The maximum speed of the car should be 13.7 m/s

Explanation:

For the car to travel at a maximum safe speed , the frictional force acting should be maximum and at the same time should provide the necessary centripetal force.

Let 'k' (=0.3502) be the coefficient of friction and 'N' be the normal force acting on the surface.

Then ,

N = mg , where 'm' is the mass of the body and 'g'(=9.8) is the acceleration due to gravity.

∴ Maximum frictional force , f = kN = kmg

Centripetal force that should act on the car to move with maximum possible speed is -

F = \frac{mv^{2} }{r}  , where 'v' is the velocity of the car and 'r'(=55m) is the radius of circular path.

Equating the 2 forces , we get -

\frac{mv^{2} }{r} = kmg

∴ v = \sqrt{krg}

Substituting all the values , we get -

v = 13.7 m/s.

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the musical note A above middle C has a frequency of 440 Hz. If the speed of the sound is known to be 350 m/s, what is the wavel
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Consider two copper wires. one has twice the length and twice the cross-sectional area of the other. how do the resistances of t
Anestetic [448]

Answer:

R = ρ L / A     where R is the resistance and ρ the resistivity:

R2 / R1 = (L2 / A2) / (L1 / A1) = L2 * A1 / (L1 * A2)

R2 / R1 =  (2 L * A) / (L * 2 A) = 1

Their resistances should be the same.

5 0
2 years ago
When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming
Zielflug [23.3K]

COMPLETE QUESTION:

<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf \varepsilon is

(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }

and since from Ohm's law

\varepsilon  = IR,

then equation (1) becomes

(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.

(a).

We are told that the change in magnetic flux is \Phi_B = 30Tm^2,  the current induced is I = 40A, and the resistance of the wire is R = 2.5\Omega; therefore, equation (2) gives

(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

which we solve for \Delta t to get:

\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},

\boxed{\Delta t = 0.3s},

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been I =20A, then equation (2) would give

(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},

\boxed{\Delta t = 0.6 s\\}

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux \dfrac{\Delta \Phi_B}{\Delta t} is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0
3 years ago
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