All categories

2 years ago

PLEASE HELP I NEED HELP PERIODTT

Physics

2 answers:

juin [17]2 years ago

7 0

Radiation PERIOD LOL
...

SVETLANKA909090 [29]2 years ago

5 0

Answer:

Explanation:

Liluzivert

You might be interested in

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

2 years ago

Mateo and Jackie were on a playground rolling some different-sized balls to get them to crash. They used a tennis ball (the same

creativ13 [48]

Answer:

In crash one, because the pink soccer ball had more mass, so it would have a stronger force.

6 0

2 years ago

What upward gravitational force does a 5600kg elephant exert on the earth?

Anastaziya [24]

Force is determined by multiplying mass and gravity (F= mg). To determine the answer, the mass of the elephant (5600 kg) is multiplied with the gravity (9.8 m/s²). The answer is 5800 N. This is the upward gravitational force that the elephant exerts on the earth.

8 0

3 years ago

Read 2 more answers

HELP!! ALL MY POINTS WILL BE GIVEN

Marta_Voda [28]

Answer:

-6 m/s^2

Explanation:

30 - 90 = -60

-60 / 10 = -6

If acceleration was constant, it will be -6 m/s^2

5 0

1 year ago

Read 2 more answers

Suppose that the Sun started shrinking in size, without losing any mass. What would be the effect of the Sun's change on the orb

adell [148]

Answer:

F = G M m / R^2 gravitational force on planet of mass m.

None of these quantities change in the given hypothesis so

there will be no change in the orbit of mass m

3 0

1 year ago

PLEASE HELP I NEED HELP PERIODTT​

PLEASE HELP I NEED HELP PERIODTT