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3 years ago

If a rock has a mass of 0.15 kg on the moon what will its mass be on earth

Physics

1 answer:

Dafna11 [192]3 years ago

6 0

The same mass dose not change but weight does

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The waves produced on the earth’s surface is called

Maslowich

Answer:

seismic wave

Explanation:

These waves are produced because of plate tectonics
These helps are useful in measurements of earthquake.
Used in seismograph.

6 0

2 years ago

Read 2 more answers

How long will it take to travel 200 km traveling 10 m/s ?

GuDViN [60]

<span>Convert 200 km to meters which equals 200000 meters. Then take 10m/s*200000meters which equals 2000000 seconds.
</span>

5 0

4 years ago

All electric devices are required to have identifying plates that specify their electrical characteristics. The plate on a certa

masha68 [24]

Answer:

R= 20 ohm

Explanation:

Given that

Current ,I = 6 A

Voltage difference ,ΔV = 120 V

Lets take resistance of the stem iron = R

We know that ,the relationship between current ,voltage difference and resistance is given as

ΔV = I R

$R=\dfrac{\Delta V}{I}$

Now by putting the values in the above equation we get

$R=\dfrac{120}{6}\ ohm$

R= 20 ohm

Therefore the resistance of the steam will be 20 ohm.

3 0

3 years ago

Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o

kolbaska11 [484]

The question ask to calculate and choose among the following is the distance of the microphone from the left - most speaker in order to pick up the loudest sound and the best answer would be letter C. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications

4 0

3 years ago

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

3 years ago