Ask question

Ask question

All categories

3 years ago

14

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

fficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?

1 answer:

ikadub [295]3 years ago

5 0

Answer:

$f_{fr}=1590.85 N$

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

$f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r}$ (1)

f(fr) is the friction force
N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

$Ncos(25)-mg-f_{fr}sin(25)=0$ (2)

Let's combine (1) and (2) to find f(fr)

$f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=1590.85 N$

I hope it helps you!

You might be interested in

What will happen to plant height if the amount of available light is reduced due to global dimming? Which type of investigation

Lorico [155]

The plant will not grow. In fact it could have all the nutrients and all the water it needs, but without a sufficient amount of light, it could die because its leaves are meant for a certain minimum amount of light.

I'll come back and see if you have posted the question you wanted and edit my answer.

6 0

3 years ago

Which of the following improves your range of motion and helps prevent

erma4kov [3.2K]

A. A healthy body composition will improve range of motion and prevent injuries. Staying in a healthy weight range (BMI) prevents overexerting joints. It also helps increase range of motion which in turn reduces injury.

3 0

3 years ago

Carmen is helping load furniture and boxes onto a moving truck. She

drek231 [11]

I think it’s contact I’m not sure

7 0

3 years ago

Read 2 more answers

Can someone solve this problem and explain to me how you got it

Zarrin [17]

1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

$F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}$

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

$F=k\frac{q_1 q_2}{r^2}$

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

$q_1' = 8q_1$

$q_2' = 8q_2$

Substituting into the equation, we find the new force:

$F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F$

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

A sledge hammer has a mass of 4.0 kg and is moving at a speed of 6.0m/s when it strikes a fence post, driving it 10 cm further i

Afina-wow [57]

Part a)

kinetic energy of the hammer will be

$KE = \frac{1}{2} mv^2$

here we know that

m = 4 kg

v = 6 m/s

now by the equation of kinetic energy we have

$KE = \frac{1}{2}*4 * 6*6$

$KE = 72 J$

Part b)

for finding the force we will have

$v_f^2 - v_i^2 = 2 a d$

here we know that

$v_f = 0$

$v_i = 6 m/s$

$d = 10 cm = 0.10 m$

now from above equation we have

$0 - 6^2 = 2 * a * 0.10$

$a = -180 m/s^2$

now by Newton's II law we have

$F = ma$

$F = 4*180 = 720 N$

so it requires 720 N.

8 0

3 years ago