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3 years ago

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

of the circle?

1 answer:

Gnom [1K]3 years ago

7 0

Answer:

r = mv^2/ Fc

Explanation:

Got it right

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Đổi 20độC = ? độ F =? Độ k

san4es73 [151]

Answer:

IDC

Explanation:

I DON'T UNDERSTAND........

7 0

2 years ago

An object has a kinetic energy of 225 j and a momentum of magnitude 28.3 kg · m/s. (a) find the speed of the object. m/s (b) fin

Irina18 [472]

A. hope this helps ;0

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4 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

4 years ago

Which is produced in a synthesis reaction?

yulyashka [42]

Answer:

Single Compound

Explanation:

A synthesis reaction takes two or more reactants and reacts chemically to turn them all into one substance.

An example of a synthesis reaction is the reaction that takes place between sodium (Na) and chlorine (Cl) to create table salt.

3 0

3 years ago

A 1000-kg car is traveling east at 20\:m.s^{-1}20 m . s − 1 and a 1200-kg car is traveling west at 22\:m.s^{-1}\:22 m . s − 1. W

nydimaria [60]

Answer:

490,400 J

Explanation:

Mass of first car, m = 1000 kg

Mass of second car, M = 1200 kg

velocity of first car, u = 20 m/s east

velocity of second car, U = 22 m/s west

The formula for the kinetic energy is

$k = \frac{1}{2}mv^{2}$

where, m is the body and v be the velocity of the body.

Total kinetic energy is given by

$k = \frac{1}{2}mu^{2}+\frac{1}{2}MU^{2}$

$k = \frac{1}{2}\times1000\times20^{2}+\frac{1}{2}\times1200\times22^{2}$

$k = 200000 + 290400$

k = 490,400 J

Thus, the total kinetic energy of the system is 490,400 J.

3 0

3 years ago