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sergij07 [2.7K]
4 years ago
13

One beam of electrons moves at right angles to a magnetic field. the force on these electrons is 4.9 x 10-14 newtons. a second b

eam travels at the same speed, but at a 30° angle with the magnetic field. what force is on these electrons?
Physics
1 answer:
diamong [38]4 years ago
7 0
Below are the choices that can be found elsewhere:

A. (4.9 × 10-14 newtons) · tan(30°) 
<span>B. (4.9 × 10-14 newtons) · sin(30°) </span>
<span>C. (4.9 × 10-14 newtons) · cos(30°) </span>
<span>D. (4.9 × 10-14 newtons) · arctan(30°) </span>
<span>E. (4.9 × 10-14 newtons) · arccos(30°)
</span>
<span>Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)</span>
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A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave
damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance x m by the time of collision, then train B would have traveled (2881 - x) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

t_{A} = t_{B}

t_{A} is the time spent by train A

t_{B} is the time spent by train B

From,

Velocity = \frac{Distance }{Time }\\

Time = \frac{Distance}{Velocity}

Since the time spent by the two trains is equal,

Then,

\frac{Distance_{A} }{Velocity_{A} }  = \frac{Distance_{B} }{Velocity_{B} }

{Distance_{A} = x m

{Distance_{B} = 2881 - x m

{Velocity_{A} = 100 m/s

{Velocity_{B} = 136 m/s

Hence,

\frac{x}{100} = \frac{2881 - x}{136}

136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\

x≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0
3 years ago
What is the length of the shadow cast on the vertical screen by your 10.0 cm hand if it is held at an angle of θ=30.0∘ above hor
Alchen [17]

Answer:

The  length is  D  =  5 \ cm

Explanation:

From the question we are told  that

     The  length of the  hand is  l  =  10.0 \ cm

      The  angle at the hand is  held is  \theta  =  30 ^o

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

             D  =  l * sin(\theta )

substituting values

             D  =  10 * sin (30)

             D  =  5 \ cm

5 0
4 years ago
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