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svetoff [14.1K]
3 years ago
10

Which constants and directions always apply when using gravity?​

Physics
2 answers:
Lostsunrise [7]3 years ago
6 0

Answer:

.

Explanation:

Romashka-Z-Leto [24]3 years ago
3 0

Answer:

Explanation::

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Help me pls 13 points
mezya [45]

Answer:

wave frequency

Explanation:

i took the test, trust me

7 0
2 years ago
Radio waves just like light waves can be reflected refracted and diffracted and polarized.
hoa [83]
<span>Radio waves just like light waves can be reflected refracted and diffracted and polarized.  The answer is True. </span>These characteristics are the common phenomena for electromagnetic (EM)  waves, and Radio Waves are electromagnetic Waves so much so that they obey reflection, refraction, and diffraction. 
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3 years ago
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An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal
borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be <u>80cm</u>.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is <u>20 cm.</u>

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u>I hope this helps!</u></em>

8 0
2 years ago
Read 2 more answers
The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm
Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

8 0
3 years ago
Equal masses are suspended from two separate wires made of the same material. The wires have identical lengths. The first wire h
choli [55]
D. I think is the correct answer
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3 years ago
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