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Aloiza [94]
3 years ago
15

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

o the air in the cloud and any collisions with air molecules. (a) What acceleration does the maximum electric field produce for protons in the cloud? Express your answer in SI units and as a fraction of g. (b) If the electric field remains constant, how far will the proton have to travel to reach 10% of the speed of light ( 3.00 × 10 8 m/s) if it started with negligible speed? (c) Can you neglect the effects of gravity? Explain your answer.
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

23.4843749996 m

Yes

Explanation:

E = Electric field = 2\times 10^5\ N/C

c = Speed of light = 3\times 10^8\ m/s

m = Mass of proton= 1.67\times 10^{-27}\ kg

q = Charge of electron = 1.6\times 10^{-19}\ C

Acceleration is given by

a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g

The acceleration is 1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

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2 years ago
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far
zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=\frac{N2}{mg}*l * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= \frac{241.75}{69.1*9.8} *l * tan 58

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3 0
3 years ago
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if 100 kilojoules of energy is used to heat 500 grams of water what is the temperature change of the water​
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Answer:

47.8 °C

Explanation:

Use the heat equation:

q = mCΔT

where q is the heat absorbed/lost,

m is the mass of water,

C is the specific heat capacity,

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Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.

100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT

ΔT = 47.8 °C

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3 years ago
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i
Stels [109]

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

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3 years ago
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