All categories

3 years ago

For a neutral object, how many electrons must transfer to make a net charge of -2?

Physics

1 answer:

Aliun [14]3 years ago

4 0

It must gain 2 electrons

You might be interested in

Can inserting a resistor in a circuit produce an effect similar to a short circuit?

tresset_1 [31]

Answer:

By ‘inserting’ you means to putting a resistor in series. In this case, no, there is no resistance that would produce the same effect as a short circuit.

If you adding a resistor in parallel with the circuit, then if it had a low value It might be similar to a short circuit. I

6 0

2 years ago

Please help on this one somebody? :)

ahrayia [7]

ans D. Substance B has a greater latent heat of vaporization than substance A.

8 0

3 years ago

What's the thrust force of 20,000N

nika2105 [10]

Answer:

895g

Explanation:

5 0

3 years ago

Imagine a 10kg block moving with a velocity of 20m/s to the left.

lara31 [8.8K]

Kinetic energy , KE= [1/2]m*v^2

m = 10 kg
v=20m/s

KE = [1/][(10kg)(20m/s)^2 = [1/2](10kg)(400m^2/s^2) = 2000 joule

Answer: 2000 joule

3 0

3 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago