What is true of the moon's orbital and rotational periods?

1 answer:

4 0

The moon's orbital and rotational periods are identical or the same, I<span>ts rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.</span>

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Near the equator, the Earth's magnetic field points almost horizontally to the north and has magnitude B=.5 x 10^-4T. What shoul

Nataly [62]

Base in your question about the magnetic field of the Earth near the equator where as its almost horizontally to the north and has magnitude of B=0.5x10^-4t, the answer is <span>Velocity of electron will be westwards.</span>

6 0

4 years ago

A car is traveling at 10 m/s. 10 seconds later the car is traveling 40 m/s. What is the car’s acceleration?

ratelena [41]

Answer:

a = 3 m/s^2

Explanation:

Vi = 10 m/s

Vf = 40 m/s

t = 10 s

Plug those values into the following equation:

Vf = Vi + at

40 = 10 + 10a

---> a = 3 m/s^2

3 0

3 years ago

Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.

Anika [276]

Answer : The significant digit is 6

Explanation :

Multiply $5.036\times10^{2}m$ by $0.078\times10^{-1}m$

Now, on multiplying

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}$

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}$

Now, the significant digit is 6.

Hence, this is the required solution.

3 0

3 years ago

Can somebody help me please

Eddi Din [679]

51 inches.

This is because a stem-plot is formatted as so:

If it’s 5 on the left side of the line, anything on the right is the ones place making possible numbers 51, 53, 56, etc.

Hope this helps!

3 0

3 years ago

A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$