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4 years ago

Does centripetal force change in uniform circular motion?

Physics

1 answer:

HACTEHA [7]4 years ago

7 0

It depends on how fast you are going and in orincipal no

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My older brother received a ticket for driving 80 mph this information describes my brothers?

Sauron [17]

Answer: 80mph describe the speed of your brother

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3 years ago

if you toss a stick into the air, it appears to wobble all over the place. Specifically, about what place does it wobble?

12345 [234]

Foolish, ambiguous question, with no correct answer, but at a wild guess, it MIGHT wobble about it own axis????

3 0

3 years ago

WILL GIVE BRAINLIEST ASAP

Sedaia [141]

Explanation:

1. Height Relatives to reference point, Mass, and strength of the gravitational field it's in

2. Distance in the magnetic field

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3 years ago

To win the game, a place kicker must kick a

masya89 [10]

Answer:

0.57 m

Explanation:

First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is

$v_x = u cos \theta = (15)(cos 51.7^{\circ})=9.30 m/s$

And the distance to cover is

d = 19 m

So the time taken is

$t=\frac{d}{v_x}=\frac{19}{9.30}=2.04 s$

Now we want to find how high the ball is at that time. The initial vertical velocity is

$u_y = u sin \theta = (15)(sin 51.7^{\circ})=11.77 m/s$

So the vertical position of the ball at time t is

$y(t) = u_y t - \frac{1}{2}gt^2$

where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find

$y=(11.77)(2.04)-\frac{1}{2}(9.8)(2.04)^2=3.62 m$

The crossbar height is 3.05 m, so the difference is

$\Delta h = 3.62 - 3.05 =0.57 m$

So the ball passes 0.57 m above the crossbar.

8 0

3 years ago

A 2500‐kg vehicle traveling at 25 m/s can be stopped by gently applying the breaks for 20 seconds. What is the average force sup

katen-ka-za [31]

Momentum = (mass) x (speed)

Change in momentum = (force) x (time)

The initial momentum is (mass) x (speed) = 2500x 25 = 62,500 kg-m/s.

Since you want to stop the vehicle, that number is also the required change
in momentum ... you want the vehicle to wind up with zero momentum.

62,500 = (force) x (time) = 20 x force

Divide each side by 20 :

force = 62,500 / 20 = 3,125 newtons 

3 0

3 years ago