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rewona [7]
3 years ago
8

What is a true statement about the rotation period of the moon?

Physics
1 answer:
mrs_skeptik [129]3 years ago
5 0
The time it takes for the Moon to rotate once around its axis is equal to the time it takes for the Moon to orbit once around Earth
You might be interested in
Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT
Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\\ \tt\bull\leadsto PV

\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]

\\ \tt\bull\leadsto [ML^5T^{-2}]

RHS

\\ \tt\bull\leadsto nRT

\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]

\\ \tt\bull\leadsto [ML^5T^{-2}]

LHS=RHS

hence verified

6 0
3 years ago
Determine the kinetic energy of 1000-kg roller coaster car that is moving with speed of 20.0m/s
nevsk [136]
B, i got the same question
7 0
3 years ago
A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i
evablogger [386]

Answer:

The period of motion  of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

k = \frac{F}{x}

k = \frac{0.08829}{0.035}

k = 2.522 \frac{N}{m}

New mass(m_1) = 26 gm = 0.026 kg

Now the period of motion is given by

T = 2 \pi \sqrt{\frac{m}{k} }

T = 2 \pi \sqrt{\frac{0.026}{2.522} }

T = 0.637 sec

This is the period of motion  of new mass.

3 0
3 years ago
2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotatin
ycow [4]

Answer:

\frac{0.065}{r}

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

v=\sqrt{\mu gr}....................(1)

where \mu is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = 9.81m/s^2

r = ?

\mu=?

In order to solve for \mu, we can simply make it the subject of formula from equation (1) as follows;

v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

\mu=\frac{0.8^2}{9.81r}

\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}

8 0
3 years ago
Please help, and show steps. Thank you very much!
Vikentia [17]
V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
6 0
4 years ago
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