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What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static

JulijaS [17]

Answer:

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

$U*m*g=\frac{mv^2}{r}$

$U*g=\frac{v^2}{r}$

substituting our given data in to expression we can solve for the speed V

$0.5*9.81=\frac{v^2}{94}$

making v the subject of formula we have

$0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47$

v= 21.47m/s

<em><u>hence the maximum velocity of the car is 21.47m/s</u></em>

7 0

3 years ago

Calculate the reading on voltmeter v²

Yuliya22 [10]

The reading of the voltmeter can be determined by finding the potential difference across the 2Ω resistance by using the value of current in the circuit. V=IR, here V is the potential difference across a resistance R through which a current I is flowing.

3 0

2 years ago

Read 2 more answers

Describe metallurgy. Check all that apply.

mylen [45]

Answer:

A, C, D

Explanation:

got it right

7 0

3 years ago

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$