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lilavasa [31]
2 years ago
12

An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o

f the circle?.
Physics
1 answer:
borishaifa [10]2 years ago
7 0

Hi there!

Recall the equation for centripetal force:
F_c = \frac{mv^2}{r}

We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
r * F_c = mv^2

Divide both sides by Fc:
\boxed{ r= \frac{mv^2}{F_c}}

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A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstr
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Answer:

Explanation:

Given that

g=9.8m/s²

The spring constant is

k=50N/m

The length of the bungee cord is

Lo=32m

Height of bridge which one end of the bungee is tied is 91m

A steel ball of mass 92kg is attached to the other end of the bungee.

The potential energy(Us) of the steel ball before dropped from the bridge is given as

P.E= mgh

P.E= 92×9.8×91

P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

Uc= ½ke²

Where 'e' is the extension

Then the extension is final height extended by cord minus height of cord

e=hf - hi

e=hf - 32

Uc= ½×50×(hf-32)²

Uc=25(hf-32)²

Using conservation of energy,

Then,

The potential energy of free fall equals the potential energy in string

Uc=Us

25(hf-32)²=82045.6

(hf-32)² = 82045.6/25

(hf-32)²=3281.825

Take square root of both sides

√(hf-32)²=√(3281.825)

hf-32=57.29

hf=57.29+32

hf=89.29m

We neglect the negative sign of the root because the string cannot compressed

3 0
3 years ago
A compact car, mass 660 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the bra
Maslowich

Answer:59.43\times 10^3 kg-m/s

Explanation:

Given

mass of car m=660 kg

Initial velocity of car =102 km/h\approx 28.33 m/s towards east

Time taken to stop t=2.1 s

Force exerted F_{avg}=4.8\times 10^3 N

change in momentum is given  by impulse imparted to the car

Impulse(J)=-F\cdot t

J=-4.8\times 10^3\times 2.1

J=-59.49\times 10^3 kg-m/s

negative Sign indicates that impulse is imparted opposite to the direction of motion

magnitude of momentum J=59.49\times 10^3

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