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3 years ago

11

How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at

a speed of 23.5 m/s?

1 answer:

tatuchka [14]3 years ago

4 0

Answer:

0.66

Explanation:

By using the formula

u = v^2 / r g

Where u is coefficent of friction

u = 23.5 × 23.5 / (85 × 9.8)

u = 0.66

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Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.

Strike441 [17]

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

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2 years ago

Without fluid friction, all objects accelerate at?

lys-0071 [83]

Answer:

Acceleration of gravity= $9.8m/s/s$

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2 years ago

Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of

steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

$I = I_0 cos^2\theta$

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Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

$\frac{I}{I_0} = \frac{1}{2}$

Using the Malus Law we have,

$I = I_0 cos^2\theta$

$cos^2\theta = \frac{I}{I_0}$

$cos^2\theta = \frac{1}{2}$

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2 years ago

block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally

Alexeev081 [22]

Answer:

$v_{2} =\frac{1}{2}$

Explanation:

From the second law of Newton movement laws, we have:

$F=m*a$ , and we know that a is the acceleration, which definition is:

$a=\frac{dv}{dt}$ , so:

$F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}$

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

$dv=\frac{1}{8}(t+1)dt\\\int\limits^{v_{2}}_0 \, dv=\int\limits^{2}_{0} {\frac{1}{8}(t+1)} \, dt$

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3 years ago

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Free_Kalibri [48]

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3 years ago