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zalisa [80]
3 years ago
12

The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this fiel

d? (Enter the magnitude.) m/s2
Physics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer:

Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

charge of electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

Force  = mass x acceleration

6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

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7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
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Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

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3 0
3 years ago
When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106J/kg and the boiling point is 120∘C. At th
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Answer:

2033219.05 J

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V = Volume

P = Pressure = 2 atm

m = Mass of water = 1 kg

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W=-P\Delta V\\\Rightarrow W=-2\times 101325\times (1\times 10^{-3}-0.824)\\\Rightarrow W=166780.95\ J

Work done is 166780.95 J

Change in internal energy is given by

\Delta U=Q-W

Heat is given by

Q=mL_v\\\Rightarrow Q=1\times 2.2\times 10^6\\\Rightarrow Q=2.2\times 10^6\ J

\Delta U=2.2\times 10^6-166780.95\\\Rightarrow \Delta U=2033219.05\ J

The increase in internal energy of the water is 2033219.05 J

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How many turns are needed in a solenoid of radius 10 cm and length 20 cm for its self-inductance to be 6.0 H?
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Answer:

B

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