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amid [387]
3 years ago
12

An electron is moving at 7.4 x 10^5 m/s perpendicular to a magnetic field. It experiences a force of -2.0 x 10^-13 N. What is th

e magnetic field strength?
Physics
1 answer:
irina1246 [14]3 years ago
6 0

Answer: B = 1.69 T

Explanation: Given that the

Electron velocity V = 7.4 × 10^5 m/s

Force F = - 2 × 10^-13 N

Charge q =  -1.602 x 10-19 C.

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation F = qvB sin θ .

Since the electron is moving perpendicular to a magnetic field,

sin θ = sin 90 = 1, therefore

- 2 × 10^-13 = - 1.602×10^-19 × 7.4×10^5B

B = -2×10^-13/1.2×10^-13

B = 1.68708T

B = 1.69 T

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A circuit with two or more branches for current to flow
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If there's any point in a circuit where the current has a choice
of which branch to take, then you have a <em>parallel circuit</em>.


4 0
3 years ago
an object is thrown with an initial horizontal velocity of 10 meters per second and take approximately 9 seconds to reach the gr
pantera1 [17]

The horizontal velocity was constant, so:

s = vt

s = 10\cdot9

s = 90

it traveled 90meters

6 0
3 years ago
A. An automobile mass is 3.5 x103 kg. If the forward thrust (Fnet)
katrin [286]

Answer:

a = 0.8 m/s^2

Explanation:

Force equation: F = ma

F = ma -> a = F/m = 2.8*10^3 N / 3.5*10^3 kg = 0.8 m/s^2

8 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
What does every magnet possess?
IRINA_888 [86]

Answer:

a North and South Pole :)

Explanation:

5 0
3 years ago
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