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3 years ago

5

How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius

90 m at a speed of 90 km/h ?

1 answer:

Korvikt [17]3 years ago

4 0

Answer:

<h2>The coefficient of static friction will be 0.7</h2>

Explanation:

Given data

the radius of curve= 90m

speed v= 90 km/h to m/s = (90*100)/60*60= 25 m/s

we know that the expression for the centripetal force acting on the car

$Fc= \frac{mv^2}{r}$ -------1

we also know that the expression for the frictional force between road and tire.

Ff= μmg--------2

Equating equation 1 and 2 we have

μmg= mv^2/r

μ= v^2/gr

substituting the values of speed and radius we have (assuming g= 9.81m/s^2)

μ= 25^2/9.81*90

μ= 625/882.9

μ= 0.7

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GarryVolchara [31]

Answer:

Part a)

$y = 88.5 m$

Part b)

$v_x = 7.7 m/s$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

$x = 30.8 m$

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 2.5(4) + \frac{1}{2}(9.81)(4^2)$

$y = 88.5 m$

Part b)

components of the velocity is given as

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

distance where it will strike the floor is given as

$x = v_x t$

$x = 7.7 \times 4$

$x = 30.8 m$

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

$y = v_y t + \frac{1}{2}gt^2$

$10 = 2.5t + \frac{1}{2}(9.81) t^2$

t = 1.2 s

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3 years ago

A golfer strikes a 0.059-kg golf ball with a force of 290 N. If the ball moves with a velocity of 69 m/s,

Ymorist [56]

Answer:

0.014s

Explanation:

Given parameters:

mass of golf ball = 0.059kg

force applied = 290N

velocity = 69m/s

initial velocity = 0m/s

Unknown:

Time of contact = ?

Solution;

We know that momentum is the quantity of motion of body possess;

Momentum = mass x velocity

Momentum = 0.059 x 69 = 4.1kgm/s

Also; impulse is the effect of the force acting on a body;

impulse = force x time = momentum

So;

Force x time = momentum

Time = $\frac{momentum}{force }$ = $\frac{4.1}{290}$ = 0.014s

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3 years ago

This problem has been solved!

irakobra [83]

Answer:

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Explanation:

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Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the

Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

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The energy equation is:

$q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +delta-p$

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

$0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +0$

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

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Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

$m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}$

The exit area is:

$m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}$

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3 years ago

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Anna007 [38]

Answer: decantation and screening

Explanation:

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