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love history [14]

3 years ago

7

A thin conducting square plate 1.0 m on the side is given a charge of-2.0 x 10-6 c. A proton is placed 1.0 en above the center o

f the plate, what is the acceleration of the proton? (Enter the magnitude in m/s.) magnitude direction Select the plate m/s2

1 answer:

adell [148]3 years ago

3 0

Answer:

Acceleration, $a=1.08\times 10^{13}\ m/s^2$

Explanation:

It is given that,

Side of the square plate, l = 1 m

Charge on the square plate, $Q=-2\times 10^{-6}\ C$

Position of a proton, x = 1 cm

The electric field due to a parallel plate is given by :

$E=\dfrac{Q}{2A\epsilon_o}$

Electric force is given by :

F = q E

$F=\dfrac{Qe}{2A\epsilon_o}$

e is the charge on electron

The acceleration of the proton can be calculated as :

$a=\dfrac{F}{m}$

m is the mass of proton

$a=\dfrac{Qe}{2A\epsilon_o m}$

$a=\dfrac{2\times 10^{-6}\times 1.6\times 10^{-19}}{2(1)^2\times 8.85\times 10^{-12}\times {1.67\times 10^{-27}}}$

$a=1.08\times 10^{13}\ m/s^2$

So, the acceleration of the proton is $1.08\times 10^{13}\ m/s^2$ . Hence, this is the required solution.

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faust18 [17]

Answer:

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2 years ago

If a train is travelling 200km/hour eastward for 1800 seconds how far does it travel?

Olenka [21]

Answer:

Distance, d = 99990 meters

Explanation:

It is given that,

Speed of the train, v = 200 km/h = 55.55 m/s

Time taken, t = 1800 s

Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :

$v=\dfrac{d}{t}$

$d=v\times t$

$d=55.55\times 1800$

d = 99990 m

So, the distance covered by the train is 99990 meters. Hence, this is the required solution.

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2 years ago

The current estimate for the age of earth of 4.57 ga comes from _____.

Whitepunk [10]

The estimation of the age of the Earth comes from scientists who studied the rock formation under the Earth's crust. They believe that during the Big Bang, the Earth, together with the other celestial bodies in the universe, were formed. Under the Earth's soil contain radioactive materials that have known decay rates. Using these rate, they undergo tests like rock dating to find the time of its existence.

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2 years ago

A calorimeter contained 350.0 g of water [cp=4.18 J/(g °C)] at 24.0 °C. An electric current was passed through a heater placed i

Shtirlitz [24]

Cp shows the amount of energy needed to raise temperature by one degree for one gram of water.

Formula for calculating cp is:
$cp= \frac{energy}{(mass)*( temperature_{change} ))} \\ temperature_{change}= \frac{energy}{(mass)*( cp))} \\ \\ temperature_{change}= \frac{16700}{(350)*( 4.18))} \\ \\ temperature_{change}=2.73 \\ \\ temperature_{final} =temperature_{initial}+temperature_{change} \\ temperature_{final}=24 + 2.73 \\ temperature_{final}=26.73$

Final temperature is 26.73°C.

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3 years ago

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in t

Liono4ka [1.6K]

(a) 5.69 N/C, vertically downward

We can calculate the acceleration of the electron by using the SUVAT equation:

$d=ut+\frac{1}{2}at^2$

where

d = 4.50 m is the distance travelled by the electron

u = 0 is the initial velocity of the electron

$t=3.00 \mu s = 3.0 \cdot 10^{-6} s$ is the time of travelling

a is the acceleration

Solving for a,

$a=\frac{2d}{t^2}=\frac{2(4.50)}{(3.0\cdot 10^{-6})^2}=1.0\cdot 10^{12} m/s^2$

Given the mass of the electron,

$m=9.11\cdot 10^{-31} kg$

We can find the electric force acting on the electron:

$F=ma=(9.11\cdot 10^{-31})(1.0\cdot 10^{12})=9.11\cdot 10^{-19}N$

And the electric force can be written as

$F=qE$

where

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

E is the magnitude of the electric field

Solving for E,

$E=\frac{F}{q}=\frac{9.11\cdot 10^{-19}}{-1.6\cdot 10^{-19}}=-5.69 N/C$

The negative sign means that the direction of the electric field is opposite to the direction of the force (because the charge is negative): since the force has same direction of the acceleration (vertically upward), the electric field must point vertically downward.

(b) Yes

We can answer the question by calculating the magnitude of the gravitational force acting on the electron, to check if it is relevant or not. The gravitational force on the electron is:

$F=mg$

where

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.81 m/s^2$ is the acceleration due to gravity

Substituting,

$F=(9.11\cdot 10^{-31})(9.81)=8.93\cdot 10^{-30}N$

We see that the gravitational force is basically negligible compared to the electric force calculated in part (a), therefore we can say it is justified to ignore the effect of gravity in the problem.

7 0

3 years ago