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3 years ago

A person will feel weightless whenever?

2 answers:

6 0

There are two ways a person could feel weight less. Either when he is a place with zero gravity(space) or when he is In a free fall

gavmur [86]3 years ago

5 0

When your'e in space you feel weightless blc there's no grsvity in space.

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A person drives north 3 blocks, then turns east and drives 3 blocks. The driver then turns south and drives 3 blocks. How could

KonstantinChe [14]

Answer:by driving east 3 blocks from the starting point

Explanation:)

4 0

3 years ago

An object is allowed to fall freely near the surface of an unknown planet. The object falls 72 meters from rest in 4.0 seconds.

dimaraw [331]

<span>it fairly is going to attain a speed of 24 m/s in a 2d, yet between t = 0 and t = a million, it fairly is not any longer vacationing at that speed, yet at slower speeds. it fairly is 12 meters. ?D = [ ( a?T^2 + 2?Tv_i ) ] / 2 the place: ?D = displacement a = acceleration ?T = elapsed time v_i = preliminary speed ?D = [ ( 24m/s^2 • 1s • 1s + 2 • 1s • 0m/s ) ] / 2 ?D = 24 / 2 ?D = 12m</span>

7 0

3 years ago

Read 2 more answers

Calculate the potential energy of 5.0g (0.005 kg) paper airplane 5.0 meters above the ground

Readme [11.4K]

PE = (mass) (gravity) (height)

PE = (0.005 kg) (9.8 m/s²) (5 m)

<em>PE = 0.245 Joule</em>

4 0

3 years ago

A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo

guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}$

If we consider than the direction of the electric field is $\vec{E}=E_0\hat{x}$ , we can solve the problem differentiating the integral for each face of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6$

E₀ is a constant and each surface is equal to each other, so: $S_1=S_2=S_i=S$

Therefore:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c$

3 0

3 years ago

1 example of Deionization

xxTIMURxx [149]

Answer:

For example, the rain water does not have any salinity, except the small quantity due to the atmosphere, while the sea water has a very high salinity.

hope this helps you friend

7 0

3 years ago