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Liula [17]
3 years ago
9

What is the force, in units of femtoNewtons, on an electron moving at a velocity of 7.26 x 106 ms-1 perpendicularly to a magneti

c field when the magnetic flux density of 0.049 T?

Physics
1 answer:
Sever21 [200]3 years ago
7 0

Answer:

56.9 fN

Explanation:

The force exerted on the electron is:

F=Bqv

where

B is the magnetic field strength

q is the electron's charged

v is the speed of the electron

In our problem,

B = 0.049 T

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

v=7.26\cdot 10^6 m/s

Therefore, the force is

F=(0.049 T)(1.6\cdot 10^{-19} C)(7.26\cdot 10^6 m/s)=5.69\cdot 10^{-14} N=56.9 fN

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W = Qin  -  Qout
     = 0.5 x 10⁵ J

By definition, the efficiency is
η = W/Qin
   = 100*(0.5 x 10⁵/4 x 10⁵)
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Answer: The efficiency is 12.5%
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What will happen to plant height if the amount of available light is reduced due to global dimming? Which type of investigation
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A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 \frac{m}{s^{2} }

Wc= 2230*9.8

Wc=21854 N

Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

N=W

N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

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8 0
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There are several information's of immense importance already given in the question. Based on the given information's the answer to the question can easily be determined.
Distance covered by the bicycle = 5000 meter
Time taken by the bicycle to reach the distance = 500 second.
Velocity of the bicycle = Distance / Time taken
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So the velocity of the bicycle is 50 meter per second. I hope the procedure is clear enough for you to understand. In future you can always use this procedure for solving similar problems.
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