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Ierofanga [76]
3 years ago
8

An electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen

and directed toward the right. What is the direction of the magnetic force on the electron?
Physics
1 answer:
Ad libitum [116K]3 years ago
6 0

Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.

Explanation:

The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:

F = BqVSinØ

If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field

According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.

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If s is the specific heat capacity of 4kg water, what is the specific heat capacity of 16kg
uranmaximum [27]

Answer: d

Explanation: divide it by 4

6 0
2 years ago
A space traveler discovers that her weight on a new planet is 192 newtons. Her mass is 68 kilograms. What is the gravitational a
alina1380 [7]

Answer:

g ≈ 2.82 m/s^2

Explanation:

By W = mg,

W = weight (in newtons)

m = mass (in kg)

g = gravitational acceleration (in m/s^2)

192 = 68g

g = 2.82352941176 m/s^2

g ≈ 2.82 m/s^2

3 0
2 years ago
A child is riding a merry-go-round that is turning at 7.18 rpm. If the child is standing 4.65 m from the center of the merry-go-
saul85 [17]

Answer:

Hello! It should be 3.50m/s! Hope it helps.

Explanation:

4 0
2 years ago
Problem 02.061 For the given circuit, assume vS = 10 V, R1 = 9 Ω, R2 = 4 Ω, R3 = 4 Ω, R4 = 5 Ω, and R5 = 4 Ω. Reference Book &am
stich3 [128]

Answer:

Ws=8.75 Watts

Explanation:

As per fig. of prob 02.061, it is clear that R5 and R4 are in parallel, its equivalent Resistance will be:

\frac{1}{Req45}=\frac{1}{R4}+\frac{1}{R5}

\frac{1}{Req45}=\frac{1}{5}+\frac{1}{4}

\frac{1}{Req45}=0.2+0.25=0.45\\ Req45=2.22

Now, this equivalent Req45 is in series with R3, therefore:

Req345=R3+Req45\\Req345=4+2.22\\Req345=6.22

This Req345 is in parallel with R2, i.e

Req2345=(R2^{-1}+Req345^{-1}  )^{-1}\\ Req2345=(4^{-1}+6.22^{-1}  )^{-1} \\Req2345=2.43

Now this gets in series with R1:

Req12345=R1+Req2345\\Req12345=9+2.43\\Req12345=11.43

Now, the power delivered Ws is:

Ws=Vs*I=\frac{Vs^{2}}{Req}  \\Ws=\frac{10^{2} }{11.43} \\Ws=8.75 Watts

8 0
3 years ago
A small fish is dropped by a pelican that is rising steadily at 0.500 m/s. How far below the pelican is the fish after 2.50 s?
sesenic [268]
Refer to the diagram shown below.

h = original height of the pelican when the fish is dropped (not relevant).
S =  distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.

Use the following equation:
S = ut + (1/2)gt²

S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
   = -29.375 m

The negative sign means that the fish drops by  29.375 m from the original height of h.

Answer: The fish is 29.375 m below where the pelican dropped it after 2.5 s.

7 0
3 years ago
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