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Tcecarenko [31]

3 years ago

14

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

r from sliding down the incline.

1 answer:

igomit [66]3 years ago

3 0

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

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This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

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Answer:

Part a)

$F = 135.7 N$

Part b)

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Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg + F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg + \mu(Fsin\theta)$

$F(cos\theta - \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta - \mu sin\theta}$

$F = \frac{55}{cos50 - 0.310(sin50)}$

$F = 135.7 N$

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg - F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg - \mu(Fsin\theta)$

$F(cos\theta + \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta + \mu sin\theta}$

$F = \frac{55}{cos50 + 0.310(sin50)}$

$F = 62.5 N$

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Answer:

The maximum safe depth in salt water is 3758.2 m.

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Given that,

Diameter = 20 cm

Radius = 10 cm

Thickness = 9.0 cm

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Inside pressure = 1.0 atm

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

Put the value into the formula

$A=\pi\times(10\times10^{-2})^2$

$A=0.0314\ m^2$

We need to calculate the pressure

Using formula of pressure

$P=\dfrac{F}{A}$

Put the value into the formula

$P=\dfrac{1.2\times10^{6}}{0.0314}$

$P=38216560.50\ Pa$

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Using equation of pressure

$P=P_{atm}+\rho gh$

$h=\dfrac{P-P_{atm}}{\rho g}$

Put the value into the formula

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