All categories

3 years ago

How does a planet’s size affect the length of a rotation of that planet

Physics

1 answer:

svet-max [94.6K]3 years ago

5 0

There doesn't seem to be any direct connection.

You might be interested in

A crate which has a mass of m1 = 125kg is sitting on an icy surface. A rope is attached to the crate and held at angle theta 28.

devlian [24]

Answer:

μ = 0.109

Explanation:

Draw a free body diagram of the crate. There are four forces:

Weight force mg pulling down.

Normal force N pushing up.

Applied force P pulling at θ above the horizontal.

Friction force Nμ pushing to the left.

Sum of the forces in the y direction:

∑F = ma

N + P sin θ − mg = 0

N = mg − P sin θ

Sum of the forces in the x direction:

∑F = ma

P cos θ − Nμ = ma

P cos θ − ma = Nμ

μ = (P cos θ − ma) / N

μ = (P cos θ − ma) / (mg − P sin θ)

Given:

P = 585 N

θ = 28.0°

m = 125 kg

a = 3.30 m/s²

μ = (585 cos 28.0° − 125 kg × 3.30 m/s²) / (125 kg × 9.8 m/s² − 585 sin 28.0°)

μ = 0.109

3 0

3 years ago

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

s344n2d4d5 [400]

Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)

Formula (Kinetic energy)
$E_{k} = \frac{m*s^2}{2}$

Solving:
$E_{k} = \frac{m*s^2}{2}$
$E_{k} = \frac{1*3^2}{2}$
$E_{k} = \frac{1*9}{2}$
$E_{k} = \frac{9}{2}$
$\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

Which of the following is a resistor? fan switch battery Thank you!

Rama09 [41]

The answer is battery:)

6 0

3 years ago

Read 2 more answers

A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease

skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

$P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm$

So, the new pressure is 2 atm.

7 0

3 years ago

Which of the following is true of high clouds?

Wewaii [24]

I’m pretty sure it’s B

3 0

2 years ago