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mafiozo [28]
3 years ago
12

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

e electric force on this electron due to this field?
1.6 × 10-3 newtons
1.4 × 1024 newtons
2.2 × 10-14 newtons
7.4 × 10-13 newtons
4.5 × 1014 newtons
Physics
2 answers:
RSB [31]3 years ago
7 0
F=Eq. Just sub in the values and you are done.
ivann1987 [24]3 years ago
3 0

Answer:

2.2 x 10^-14 N

Explanation:

q = - 1.6 x 10^-19 C

E = 1.4 x 10^5 N/C

The force on a charged particle placed in an electric field is given by

F = q x E

F = - 1.6 x 10^-19 x 1.4 x 10^5 = - 2.24 x 10^-14 N

F = - 2.2 x 10^-14 N

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A. 131 V

I got it right on the quiz

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3 years ago
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You push a refrigerator with a force of 100 N. If you move the refrigerator a distance of 5 m while you are pushing, how much wo
ddd [48]

The work done to push the refrigerator is 500 Nm.

Explanation:

Work done is the measure of force required to move any object from one point to another. So it is calculated as the product of force and displacement.

If the force increases the work done will increase and similarly, the increase in displacement increases the work done. So to push the refrigerator work should be done on the object and not by the object.

As the force is 100 N and the displacement is 5 m then, work done can be measured as

Work = Force × Displacement

Work = 100 × 5 = 500 Nm

So the work done to push the refrigerator is 500 Nm.

7 0
3 years ago
How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t
tekilochka [14]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

Q=mC\Delta T\\\\Q=\rho VC\Delta T

(m=\rho V)

where,

Q = amount of heat required = ?

m = mass

\rho = density of air = 1.20kg/m^3

V = volume of air

C = specific heat of air = 1006J/kg^oC

\Delta T = change in temperature = 10.0^oC

Now put all the given values in above formula, we get:

Q=\rho VC\Delta T

Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)

Q=1.449\times 10^6J

Now we have to calculate the time required.

Formula used :

t=\frac{Q}{P}

where,

t = time required = ?

Q = amount of heat required = 1.449\times 10^6J

P = power = 1500 W

Now put all the given values in above formula, we get:

t=\frac{1.449\times 10^6J}{1500W}

t=966s\times \frac{1min}{60s}=16.1min

Thus, the time required is, 16.1 minutes.

5 0
3 years ago
The acceleration of a particle is given by ax(t) = -(2.00 m/s2) + (2.70 m/s3)t. (a) find the initial velocity v0x, such that the
Alexeev081 [22]
I’ll solve them right now just wait a few
3 0
3 years ago
Why can the atmosphere hold on to heat
Naddika [18.5K]

Answer:

A. Air

Explanation:

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2 years ago
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