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Lubov Fominskaja [6]
3 years ago
12

An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 1.95 mT. If the speed of the

electron is 1.44 107 m/s, determine the following. (a) the radius of the circular path _____ cm (b) the time interval required to complete one revolution ____ s
Physics
1 answer:
zvonat [6]3 years ago
4 0

Answer: a) 42 *10^-3 m= 42 mm; b) 10.40 *10^-6 s=10.40 μs

Explanation: In order to response this problem we have to consider the Newton law for the circular movement,

Fm=m*ac where Fm is the magnetci force and ac the centripetal acceleration which is equal to v^2*R ( the raduis of teh curcular trajectory)

Fm=e*v*B considering that v and B are perpendicular

then we have:

e*v*B=m*v^2/R so

R=m*v/(e*B)= 9.1*10^-31*1.44*10^7/(1.6*10^-19*1.95*10^-3)= 42 mm

Then to calculate the time to complete one revolution ( period)

we know that ω=v*R and T= 2π/ω

then we have:

ω=1.44*10^7*42*10^-3=604.8 *10^3 rad/s

T=2*π/604.8 *10^3 = 10.4 μs

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3 years ago
Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be
Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

v_1=\dfrac{72.42 }{2.5}=28.96\ km/h

Average speed of Prav,

v_2=\dfrac{119}{1.25}=95.2\ km/h

Difference in average speed of Ajoba and Prav is :

v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

7 0
3 years ago
Which of the following has mechanical energy?
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5 0
3 years ago
Read 2 more answers
A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

6 0
3 years ago
Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
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