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9966 [12]
3 years ago
12

A certain moving electron has a kinetic energy of 1.01 × 10−19 J. Calculate the speed necessary for the electron to have this en

ergy. The mass of an electron is 9.109 × 10−31 kg. Answer in units of m/s.
Physics
1 answer:
insens350 [35]3 years ago
5 0

Answer:

Velocity of electron will be equal to v=4.711\times 10^5m/sec

Explanation:

We have given kinetic energy of electron KE=1.01\times 10^{-19}J

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

We have to find velocity of electron

Kinetic energy of electron is equal to KE=\frac{1}{2}mv^2, here m is mass and v is velocity

So 1.01\times 10^{-19}=\frac{1}{2}\times 9.1\times 10^{-31}\times v^2

v=4.711\times 10^5m/sec

So velocity of electron will be equal to v=4.711\times 10^5m/sec

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A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity
Crank

Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

8 0
2 years ago
Problem 21-40a:
Greeley [361]

Answer:

Part a)

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

q = 3.37 \times 10^{-4} C

Explanation:

As we know that electric force on electric charge is given as

F = qE

here we have

q = 1.6 \times 10^{-19}C

E = 153 N/C

now force is given as

F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N

Gravitational force on electric charge near surface of earth is given as

F_g = mg

F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N

now the ratio of two forces is given as

\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

F_g = qE

mg = qE

(5.25 \times 10^{-3})(9.81) = q(153)

here we have

q = 3.37 \times 10^{-4} C

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