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3 years ago

Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.

Physics

1 answer:

Kamila [148]3 years ago

8 0

Answer: FR=2.330kN

Explanation:

Write down x and y components.

Fx= FSin30°

Fy= FCos30°

Choose the forces acting up and right as positive.

∑(FR) =∑(Fx )

(FR) x= 5-Fsin30°= 5-0.5F

(FR) y= Fcos30°-4= 0.8660-F

Use Pythagoras theorem

F2R= √F2-11.93F+41

Differentiate both sides

2FRdFR/dF= 2F- 11.93

Set dFR/dF to 0

2F= 11.93

F= 5.964kN

Substitute value back into FR

FR= √F2(F square) - 11.93F + 41

FR=√(5.964)(5.964)-11.93(5.964)+41

FR= 2.330kN

The minimum force is 2.330kN

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A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

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The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

$v=85 mph \cdot \frac{1609}{3600}=38.0 m/s$ is the speed

And solving for $\mu$ , we find the coefficient of friction required to keep the car in circular motion:

$\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196$

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3 years ago

A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.

GenaCL600 [577]

Answer:

Impulse = 80Ns

Explanation:

Given the following data;

Mass = 3kg

Force = 20N

Time = 4 seconds

To find the impulse experienced by the object;

Impulse = force * time

Impulse = 20*4

Impulse = 80Ns

Therefore, the impulse experienced by the object is 80 Newton-seconds.

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3 years ago

How much energy ( in joules ) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature ?

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Answer:

B. 17,705.1 J

Explanation:

The hear released when the mercury condenses into a liquid is given by:

$Q=m \lambda_v$

where

m = 0.06 kg is the mass of the mercury

$\lambda_v$ is the latent heat of vaporization

For mercury, the latent heat of vaporization is $\lambda_v = 296 kJ/kg$ , so the heat released during the process is:

$Q=(0.06 kg)(296 kJ/kg)=17.76 kJ = 17,760 J$

So, the closest option is

B. 17,705.1 J

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