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Harlamova29_29 [7]
3 years ago
7

An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric

field, which has magnitude E = 48 N/C. (a) What is the speed of the electron 1.3 ns after entering this region? (b) How far does the electron travel during the 1.3 ns interval?
Physics
1 answer:
8090 [49]3 years ago
3 0

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

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A person drives north 8 blocks, then turns west and drives 4 blocks. The driver then turns south and drives 8 blocks. How could
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An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 2
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Answer:

The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have

The temperature T = 602.64K and the final pressure P = 110.24MPa

Explanation:

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PV = n×R×T from where we have V = final volume thus

P = (n×R×T)/V = (3×8.134×602.64)÷(133.4×10^{-6}) = 110237041.1 N/m^{2} = 110.237MPa

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