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andreev551 [17]
3 years ago
9

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

car can safely travel if the radius of the track is 79.0 m and the coefficient of friction is 0.39?
Physics
1 answer:
ad-work [718]3 years ago
6 0

Answer:

Maximum speed of the car is 17.37 m/s.

Explanation:

Given that,

Radius of the circular track, r = 79 m

The coefficient of friction, \mu=0.39

To find,

The maximum speed of car.

Solution,

Let v is the maximum speed of the car at which it can safely travel. It can be calculated by balancing the centripetal force and the gravitational force acting on it as :

v=\sqrt{\mu rg}

v=\sqrt{0.39\times 79\times 9.8}

v = 17.37 m/s

So, the maximum speed of the car is 17.37 m/s.

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A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

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The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end
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here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

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