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vredina [299]
3 years ago
10

A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 7.1 m making 3.9 revolutions

every 9.4 second. What is the centripetal acceleration of the ball?
Physics
1 answer:
IgorC [24]3 years ago
7 0

Answer:

48,2 m/s²

Explanation:

We're gonna use the Centripetal Acceleration formula: v² / r but before that, we got to know the velocity, that is not shown clearly to us, so....

To know the velocity let's calculate the distance that the ball traveled

The circumference of a circle formula is:

2piR

2 . 3,14 . 7,1 | That is equal to 44,588 m

We know that the ball traveled this distance 3,9 times, so...

44,588 . 3,9 = 173,8932 m

Ok, now we have the distance, just need to know the time, that is 9.4 seconds.

Velocity = Distance / Time

V = 173,8932 / 9,4

V = 18,5 (approximate)

So...

We are back to the first formula:

Ca = v² / r

Ca = 18,5² / 7.1

Ca = 48,2 m/s² (approximate)

I hope it is correct, hahaha.

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Answer:

Supercells

Explanation:

supercells are rotating thunderstorms that has a well-defined radar circulation called a mesocyclone. They can sometimes produce destructive hail, severe winds, frequent lightning, and flash floods.

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What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
astraxan [27]

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

<u>F= ma</u>

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

<u>v=u +at</u>

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs


s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

<u>s= 200 m</u>


3 0
3 years ago
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Pitch is determined by the ____ of a sound wave.
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The answer is c .Frequency
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A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r =  g x μ

(6.28r)^2 / r = 9.8 x 0.1

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r = 0.02 m

6 0
3 years ago
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