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3 years ago

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Calculate the initial (from rest) acceleration of a proton in a 5.00 x 10^6 N/C electric field (such as created by a research Va

n de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

1 answer:

julsineya [31]3 years ago

5 0

Answer:

Acceleration of the proton will be equal to $4.79\times 10^{14}m/sec^2$

Explanation:

We have given electric field $E=5\times 10^6N/C$

Mass of proton is equal to $m=1.67\times 10^{-27}kg$

And charge on proton is equal to $e=1.6\times 10^{-19}C$

Electrostatic force will be responsible for the motion of proton

Electrostatic force will be equal to $F=qE=1.6\times 10^{-19}\times 5\times 10^6=8\times 10^{-13}N$

According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration

This newton force will be equal to electrostatic force

So $1.67\times 10^{-27}\times a=8\times 10^{-13}$

$a=4.79\times 10^{14}m/sec^2$

So acceleration of the proton will be equal to $4.79\times 10^{14}m/sec^2$

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Are there numbers on the y axis?

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3 years ago

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time

Ber [7]

Answer:

The value is $\Delta t = 4.0 *10^{-7} \ s$

Explanation:

From the question we are told that

The current is $\Delta I = 0.100 \ A$

The inductor is $L = 2.0mH = 2.0*10^{-3} \ H$

The voltage induced is $\epsilon = 500 V$

Generally the emf induced is mathematically represented as

$\epsilon = L * \frac{\Delta I }{\Delta t }$

Here $\Delta t$ is the time taken

=> $\Delta t = \frac{L * \Delta I }{\epsilon }$

=> $\Delta t = \frac{2*10^{-3} * 0.100 }{500 }$

=> $\Delta t = 4.0 *10^{-7} \ s$

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3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

Can anyone help me please

ArbitrLikvidat [17]

Answer:

Gravity.

Rocket ships.

Ball.

Basketball.

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Gravity has to do a lot with air. It puts the planets in there area.

Rocket Ship has to do a lot with air. If i'm right, they calculate the area, weather, about the air.

A ball gets throwed in the air, which gravity comes into place.

Basketball is also a similar example to a ball.

7 0

2 years ago

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0

3 years ago