Ask question

Ask question

All categories

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?.

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:
$\boxed{ r= \frac{mv^2}{F_c}}$

You might be interested in

One example of a physical change is

ycow [4]

Answer:

Mixing a milkshake

Explanation:

Becuse it’s physics becuse your using muscle and moving it and changing it by force.

7 0

3 years ago

What is the pressure exerted by a column of fresh water 10m high<br>(p=1000kg/m3 and g=10m/s)

alexira [117]

Height (h) = 10 m
Density (ρ) = 1000 Kg/m^3
Acceleration due to gravity (g) = 10 m/s^2
We know, pressure in a fluid = hρg
Therefore, the pressure exerted by a column of fresh water
= hρg
= (10 × 1000 × 10) Pa
= 100000 Pa

<u>Answer</u><u>:</u>

<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>

Hope you could understand.

If you have any query, feel free to ask.

8 0

3 years ago

Two pendulum bobs have equal masses and lengths (8.100 m). bob a is initially held horizontally while bob b hangs vertically at

Karo-lina-s [1.5K]

Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.

Speed will be = squreroot ( 2*g*L)

L is length of pendulum

5 0

3 years ago

Read 2 more answers

Newton’s heliocentric view of the universe meant that the sun was the center of the universe. True False

Anuta_ua [19.1K]

The answer to this question is true.

7 0

3 years ago

Read 2 more answers

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would

natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a) $v_{es} =v_{rms}$ = 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

$v_{rms} = \sqrt{ \frac{3RT}{M}}$

So, $v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}$

multiply each side by M_{o2}, so we have

$v_{rms,o2}^2 *M_{o2} =3RT_{o2}$

solving for temperature T_{o2}

$T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}$

In the question given, $v_{rms} =v_{es}$

$T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}$

T_{o2} =1.58*10^5 K

7 0

3 years ago