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3 years ago

What is the strength of an electric field that will put a force of 1.28 x 10-15 N on a proton?

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer: E = 7,490.6 N/C

Explanation:

If we have a field E, and a particle with a charge q, the force that the particle experiences is:

F = E*q

In this case, we know that the force is:

F = 1.2*10^(-15) N

And we know that the particle is a proton, where the charge of a proton is:

q = 1.602*10^(-19) C

Then we can replace these two values in the equation to get:

1.2*10^(-15) N = E*1.602*10^(-19) C

We just need to isolate E.

(1.2*10^(-15) N)/(1.602*10^(-19) C) = E

7,490.6 N/C = E

That is the strength of the electric field.

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In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to pu

tatuchka [14]

Answer:

3.90 degrees

Explanation:

Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is

W = mg = 30*9.81 = 294.3 N

This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.

We can calculate the parallel one since it's the one that affects the force required to push up

F = WsinΘ

Since customer would not complain if the force is no more than 20N

F = 20

$294.3sin\theta = 20$

$sin\theta = 20/294.3 = 0.068$

$\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0$

So the ramp cannot be larger than 3.9 degrees

6 0

3 years ago

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

-BARSIC- [3]

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62 distance traveled after collision .. crumpling of truck

Part a

$V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s$

Part b

$vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s$

7 0

4 years ago

Read 2 more answers

When you park your car uphill next to a curb, the right front wheel should be:?

Cloud [144]

When headed uphill at a curb, turn the front wheels away from the curb and let your vehicle roll backwards slowly until the rear part of the front wheel rests against the curb using it as a block.

6 0

3 years ago

1. Explain who is doing more work and why: a bricklayer carrying bricks and placing them on the wallof a building being

Basile [38]

Both are doing because they have chorus

7 0

2 years ago

For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?

V125BC [204]

Answer:

The energy of photon, $E=8\times 10^{-15}\ J$

Explanation:

It is given that,

Voltage of anode, $V=50\ kV=50\times 10^3\ V=5\times 10^4\ V$

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

$E=eV$

e is charge of electron

$E=1.6\times 10^{-19}\times 5\times 10^4$

$E=8\times 10^{-15}\ J$

So, the maximum energy of the x- ray radiation is $8\times 10^{-15}\ J$ . Hence, this is the required solution.

6 0

3 years ago