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2 years ago

If a car is rounding a flat curve on a highway, what is the centripetal force on the car?

Physics

1 answer:

tigry1 [53]2 years ago

8 0

Answer:

The net force on a car traveling around a curve is the centripetal force, Fc = m v2 / r, directed toward the center of the curve.

For a level curve, the centripetal force will be supplied by the friction force between the tires and roadway.

A banked curve can supply the centripetal force by the normal force and the weight without relying on friction.

Explanation:

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A book falls of the table and hits the floor. Is work done? Why?

morpeh [17]

<u>Yes, work is done when a book falls of the table.</u>

This is because:

When the book falls, it's potential energy is converted into kinetic energy. As it reaches the floor down, this kinetic energy is converted to heat energy and sound energy due to the impact.

When a force is imposed on an object to cause displacement of that object, work is done on that object. For a force to do work on an object, there should be a displacement and this force should cause the displacement. So here, since the book falls from the table and causes the displacement of the book from the table to the floor. It is said that work is done.

Work can be given by the formula:

W = F • d

where F is the force and d is the displacement.

6 0

2 years ago

Read 2 more answers

A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t

kirill [66]

As per momentum conservation theory

If net force on a system is zero then initial momentum of the system will be equal to final momentum of that system.

Since initially coconut is at rest so initial momentum is zero

Here it break into three pieces such that two parts of same mass while third part is of double mass

now we can say

$P_1 + P_2 + P_3 = 0$

here we know that two parts fly off with same speed v0 along south and west directions

$mv_0 (-\hat j) + mv_0 (-\hat i) + 2m \vec v = 0$

now we will have

$\vec v = \frac{v_0}{2}\hat i + \frac{v_0}{2}\hat j$

so here magnitude of the speed of the third part will be

$|v| = \sqrt{(\frac{v_0}{2})^2+(\frac{v_0}{2})^2}$

$|v| = \frac{v_0}{\sqrt2}$

and the direction of the third part will be along North East direction

7 0

2 years ago

Read 2 more answers

The highest speed of a cheetah is 100 km/hr and the highest speed of a gazelle is 80 km/hr. If at t = 0 both animals are running

Tju [1.3M]

Answer:

t=18s

Explanation:

The final position of an object moving at constant speed is given by the formula $x=x_0+vt$ , where $x_0$ is its initial position, v its speed and t the time elapsed.

For the cheetah we have $x_c=x_{0c}+v_ct$ , and for the gazelle $x_g=x_{0g}+v_gt$ . We want to know at which t their positions are equal, that is, $x_c=x_g$ , which means,

$x_{0c}+v_ct=x_{0g}+v_gt$

Where we can do:

$v_ct-v_gt=x_{0g}-x_{0c}$

$(v_c-v_g)t=x_{0g}-x_{0c}$

$t=\frac{x_{0g}-x_{0c}}{v_c-v_g}$

We then substitute the values we have (the initial position of the cheetah is 0m), writing the meters in km so distance units cancel out correctly:

$t=\frac{0.1km-0km}{100km/hr-80km/hr}=0.005hr=18s$

On the last step we just multiply by 3600 because is the number of seconds in an hour.

3 0

3 years ago

UPHILL, MORE ENGINE POWER WILL BE NEEDED TO OVERCOME THE EFFECTS OF;

posledela

Answer: Its The Force Of Gravity

I hope this helps :)

7 0

3 years ago

Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to b

weeeeeb [17]

Answer:

4.408 m/s, 4.102 m/s, 4.026 m/s

Explanation:

The question is incomplete. The text of the original question states:

A race car moves such that its position fits the relationship

$x=(4.0 m/s)t + (0.85 m/s^3) t^3$

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.

We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

$v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2$

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

$v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s$

5 0

2 years ago