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lisabon 2012 [21]
4 years ago
14

How much work does the electric field do in moving a proton from a point with a potential of +130 v to a point where it is -70 v

? express your answer in joules?
Physics
1 answer:
NikAS [45]4 years ago
6 0
Hi Gonnah,
W=qΔV
W= q(-70-130)
q of proton look it up and plug in and find out
it should be negative since you are doing work form greater potential to lower
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A car moves 8 km from home to school and then back to the home. a) Calculate the Distance covered by this car. b) What is the Di
Sveta_85 [38]

Explanation:

Distance covered = Total distance travelled by the car

Since the car travels to school and back to home again,

Total diatance covered = 8 + 8 = 16km

Displacement = Shortest distance between start point and destination.

Since, the car returns back to home from school,

the start and destination point are same.

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8 0
3 years ago
What is the frequency of an X-ray with wavelength 0.13 nm ? Assume that the wave travels in free space. Express your answer to t
dem82 [27]

Answer:

Frequency, f=2.30\times 10^{18}\ Hz

Explanation:

Given that,

The wavelength of the x-rays, \lambda=0.13\ nm=0.13\times 10^{-9}\ m

We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

c=f\lambda

f is the frequency

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.13\times 10^{-9}}\\\\f=2.30\times 10^{18}\ Hz

So, the frequency of an x-ray is 2.30\times 10^{18}\ Hz. Hence, this is the required solution.

5 0
3 years ago
An electron is accelerated from rest by a potential difference of (24.5 A) V for a distance of (4.50 B) cm. Determine the de Bro
DiKsa [7]

Answer:

206 pm

Explanation:

We are given that

Potential difference,\Delta V=24.5+A V

Distance,d=4.5+B cm

We have to determine the de Brogile wavelength of the electron.

A=11 and B=5

\Delta V=24.5+11=35.5 V

d=4.5+5=9.5 cm

Charge on electron,q =1.6\times 10^{-19} C

Mass of electron=m=9.1\times 10^{-31} kg

Speed of electron,v=\sqrt{\frac{2q\Delta V}{m}}

Using the formula

v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 35.5}{9.1\times 10^{-31}}

v=3.53\times 10^6 m/s

de Brogile wavelength, \lambda=\frac{h}{mv}

Where h=6.626\times 10^{-34}

\lambda=\frac{6.626\times 10^{-34}}{9.1\times 10^{-31}\times 3.53\times 10^6}=2.06\times 10^{-10}=206\times 10^{-12} m

1 pm=10^{-12} m

\lambda=206 pm

7 0
3 years ago
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Oxana [17]

Answer:

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Explanation:

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Compare and contrast infrasonic and ultrasonic vibrations
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Answer:

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