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Vesnalui [34]
3 years ago
5

C) 12g/cm d) 0.2g/cm

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
4 0

Explanation:

Is this ok?......I have got the answer in pascals.

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Some chlorine atoms have an atomic mass of 37, while others have an
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Need help ASAP <br> Thankss + BRAINLIST only for correct answer
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Provide the following information for each of the three types of radiations from naturally radioactive materials. Be sure to inc
scoundrel [369]

<em>Alpha radiation:</em>  Particles.  Each alpha particle is a little bundle of 2 protons and 2 neutrons ... identical to the nucleus of a Helium atom.

Charge  . . . +2 elementary charges

Mass  . . . 4 Atomic Mass Units

Relative penetrating power  . . . low

The effect an electric field would have on it . . . Since the alpha particle has a positive charge, it's repelled by other positive charges, and attracted toward negative charges.

<em>Beta radiation</em>:  Particles.  Each beta particle is an electron.

Charge  . . . -1 elementary charge

Mass  . . . 0.00055 AMU

Relative penetrating power  . . . medium

The effect an electric field would have on it . . .  repelled by other negative charges, and attracted toward positive charges.

<em>Gamma radiation</em>:  electromagnetic wave, verrrrry short wave, high frequency

Charge . . . electromagnetic wave, no charge

Mass  . . . electromagnetic wave, no mass

Relative penetrating power  . . . high

The effect an electric field would have on it . . . electromagnetic wave, no effect

5 0
3 years ago
What engine thrust is required for a rocket of mass 35 kg to leave the launching pad? 3.5 N. 35 N. 351 N. 3,500 N. 35,000 N.
soldi70 [24.7K]

In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.

Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.

The engine thrust has to be more than 343 N.

7 0
2 years ago
The distribution circuit of a residential power line is operated at 2000 V rms. This voltage must be reduced to 240 V rms for us
forsale [732]

Answer:

N_p\approx3958

Explanation:

In an ideal transformer the relationship between the voltages is proportional to the ratio between the number of turns of the windings. Thus:

\frac{V_p}{V_s} =\frac{N_p}{N_s}

Where:

V_p=Voltage\hspace{3}in\hspace{3}primary\hspace{3}coil

V_s=Voltage\hspace{3}in\hspace{3}secondary\hspace{3}coil

N_p=Turns\hspace{3}on\hspace{3}primary\hspace{3}coil

N_s=Turns\hspace{3}on\hspace{3}secondary\hspace{3}coil

So, solving for N_p

N_p=N_s*\frac{V_p}{V_s} =475*\frac{2000}{240} =3958.333333\approx3958

7 0
3 years ago
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