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statuscvo [17]
3 years ago
6

What is the Orbital Notation for Radon

Physics
1 answer:
balu736 [363]3 years ago
4 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

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A clindrical rod of uniform density is located with its center at the origin, and its axis along the axis. It rotates about its
zhenek [66]

Answer:

6093.2328 J

Explanation:

For cylindrical rod moment of inertia will be

I_X=I_Y=\frac{1}{12}m(3r^2+h^2)

I_X=I_Y=\frac{1}{12}\times 4(3\times 0.06^2+0.6^2)=0.1236 kg -m^2

we have given time =0.02 sec

Angular speed =\frac{2\pi }{T}=\frac{2\times 3.14}{0.02}=314\ rad/sec

Rotational KE = \frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.1236\times 314^2=6093.2328\ J

5 0
4 years ago
The reason a charged balloon will stick to a wall is that?
kari74 [83]
The negative charges of the balloon will stick to the positive charges on the wall.
6 0
3 years ago
A person who weighs 150 pounds on Earth would weigh ____ pounds on the moon.
enot [183]

Answer:

25lb

Explanation:

You haven't changed (you are made up of the same atoms), but the force exerted on you is different. Physicists like to say that your mass hasn't changed, only your weight.

6 0
3 years ago
How did the temperature structure of the solar nebula determine planetary composition?
STALIN [3.7K]

Explanation:

The temperature of the solar nebula was decreasing as it moved away from its center. Therefore, only heavy elements could condense in the inner solar system and terrestrial planets could not form with light elements, such as gases. In the outer solar system, the Jovian planets formed mostly with gases, since temperatures were too low to allow rocky compositions.

3 0
3 years ago
An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be
marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75  in +ve x- direction

F_1 = \dfrac{GM_1 m}{R_1^2}

F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}

F_1 = 2.019\times 10^{8}\ m

Force applied due to mass 14 Kg in -ve x-direction

F_2 = \dfrac{GM_2 m}{R_2^2}

F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}

F_2 = 0.857\times 10^{8}\ m

net force

F = F₁ + F₂

F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m

F = 1.162\times 10^{8}\ m

Using newton second law

a = \dfrac{F}{m}

a = \dfrac{ 1.162\times 10^{8}\ m}{m}

a =1.162\times 10^{8} \ m/s^2

b) As the acceleration of mass comes out to be  +ve hence, the direction will be toward the mass of 8.75 Kg

6 0
4 years ago
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