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aksik [14]
3 years ago
11

Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a dista

nce of 41.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
Physics
1 answer:
luda_lava [24]3 years ago
6 0

To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

V = \frac{kq}{r}

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}

Replacing the values we have that

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}

V_1 = 215.12V

Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}

V_2 = 328.2V

Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say

\frac{1}{2} mv^2 = q(V_2-V_1)

Here

m = mass

v = Velocity

q = Charge

V = Voltage

Rearranging to find the velocity

v = \sqrt{ \frac{2q(V_2-V_1)}{m}}

Replacing,

v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}

v = 6.3*10^6m/s

Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is 6.3*10^6m/s

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2.2 x 10-19

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When will an object dropped from rest attain a speed of 30 m/s?
stich3 [128]
<h3><u>Answer</u> :</h3>

Initial velocity = zero (i.e., free fall)

Final velocity = 30m/s

Acceleration due to gravity = 10m/s²

For a body falling freely under the action of gravity, g is taken positive.

◈ <u>First equation of kinenatics</u> :

⇒ v = u + gt

⇒ 30 = 0 + 10t

⇒ t = 30/10

⇒ <u>t = 3s</u>

Hence, object will attain a speed of 30m/s after 3s.

8 0
3 years ago
A wave has a wavelength of 125 meters is moving at a speed of 20 m/s. What is it’s frequency?
Ann [662]

Answer: 0.16Hz

Explanation:

Given that:

wavelength (λ) = 125 meters

speed (V) = 20 m/s

frequency (F) = ?

Recall that frequency is the number of cycles the wave complete in one second. And its value depends on the wavelength and speed of the wave.

So, apply the formula V = F λ

Make F the subject formula

F = V / λ

F = 20 m/s / 125 meters

F = 0.16 Hz

Thus, the frequency of the wave is 0.16 Hertz.

4 0
3 years ago
A box-shaped wood stove has dimensions of 0.75 mx 1.2 m x 0.4 m, an
Tresset [83]

Answer:

8.4 kW

Explanation:

Using the Stefan-Boltzmann law,

P = εAσT4

Where:  

P: Radiation Energy

ε: Emissivity of the Surface. Check emissivity table below of common materials.

A: Surface Area, in m^2.

σ: Stefan-Boltzmann Constant, σ=5.67 × 10-8 W/m2•K4

T:  Temperature

Plugging in values,

P = 0.85 x 3.328 x 5.67 x 10^(-8) x 205

P = 8383 W or 8.4 kW

3 0
3 years ago
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