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3 years ago

Q¹=0,07Mc Q²=2C r=1,08 cm F=.......?

1 answer:

5 0

The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = (k * q1 * q2) / (r^2) where:
q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9 N. m2/ C2
r is the distance between the two charges.

Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9 x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

4 0

3 years ago

What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a

Elena L [17]

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

6 0

3 years ago

Which year was pluto no longer considered a planet?.

katrin2010 [14]

2006, i hope this helps

6 0

2 years ago

An object with a 25 µC charge is 0.54 m away from a second charged object. They experience a force of 3250 N. What is the charge

nekit [7.7K]

25 uc charge is 0.54 m away

6 0

2 years ago

I am trying to find the magnitude of a resultant vector. Do i take inconsideration the negatives when i find the x & y compo

attashe74 [19]

Absolutely ! If you have two vectors with equal magnitudes and opposite
directions, then one of them is the negative of the other. Their correct
vector sum is zero, and that's exactly the magnitude of the resultant vector.

(Think of fifty football players pulling on each end of the rope in a tug-of-war.
Their forces are equal in magnitude but opposite in sign, and the flag that
hangs from the middle of the rope goes nowhere, because the resultant
force on it is zero.)

This gross, messy explanation is completely applicable when you're totaling up
the x-components or the y-components.

4 0

4 years ago