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aleksley [76]
2 years ago
7

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

lectron is 4.55×105 {\rm m/s} pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.355 {\rm m} east of point A?
A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.95×104 {\rm m/s}, again pointed towards the east. What is the speed of the proton at point B?
Physics
1 answer:
zimovet [89]2 years ago
3 0

Answer:

a) v = 6.25\times 10^{5} m/s

b) v = 1.73\times 10^{4} m/s

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is 4.55\times 10^5 m/s

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

a =\frac{q E}{m}

a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}

a = 2.58\times 10^{11} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

      = 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355

v = 6.25\times 10^{5} m/s

b) for proton

a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}

a = -1.41\times 10^{8} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

       = 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355

v = 1.73\times 10^{4} m/s

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A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1 ° from the normal to the plane of the l
umka2103 [35]

Answer:

Magnetic flux through the loop is 1.03 T m²

Explanation:

Given:

Magnetic field, B = 4.35 T

Radius of the circular loop, r = 0.280 m

Angle between circular loop and magnetic field, θ = 15.1⁰

Magnetic flux is determine by the relation:

\Phi = BA\cos \theta   ....(1)

Here A represents area of the circular loop.

Area of circular loop, A = πr²

Hence, the equation (1) becomes:

\Phi=B\pi r^{2} \cos \theta

Substitute the suitable values in the above equation.

\Phi=4.35\times\pi (0.28)^{2} \cos 15.1

\Phi = 1.03 T m²

8 0
2 years ago
How much heat has to be added to 508 g of copper at 22.3°C to raise the temperature of the copper to 49.8°C? (The specific heat
viva [34]

Answer:

Q = 5267J

Explanation:

Specific heat capacity of copper (S) = 0.377 J/g·°C.

Q = MSΔT

ΔT = T2 - T1

ΔT=49.8 - 22.3 = 27.5C

Q = change in energy = ?

M = mass of substance =508g

Q = (508g) * (0.377 J/g·°C) * (27.5C)

Q= 5266.69J

Approximately, Q = 5267J

8 0
2 years ago
An object initially at rest experiences an acceleration of 9.8 m/s2. How much time will it take to achieve a velocity of 58 m/s?
frez [133]
5.91(approx) seconds just divide velocity by acceleration
4 0
3 years ago
Relationship between voltage, resistance and current in a circuit
Lena [83]

Answer:

V = I×R

where -

V = potential difference across

I = current flowing in the circuit

R = Equivalent Resistance in the circuit

8 0
2 years ago
The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
likoan [24]

Answer:

14869817.395 m

Explanation:

\theta=22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians

From Rayleigh Criterion

\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0
3 years ago
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