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Marrrta [24]
3 years ago
12

This problem has been solved! See the answer A proton with a speed of 3.5x10^6 m/s is shot into a region between two plates that

are separated by distance of 0.23 m. A magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field so the proton just misses colliding with the opposite plate?
Physics
1 answer:
DIA [1.3K]3 years ago
7 0

Answer:

The magnitude of the magnetic field 'B' is 0.16 Tesla.

Explanation:

The magnitude of the magnetic field 'B' can be determined by;

             B = \frac{mV}{qR}

where: m is the mass of proton, V is its speed , q is the charge of proton and R is the distance between the plates.

Given that: speed 'V' of the proton = 3.5 × 10^{6} ms^{-2}, distance 'R' between the plates = 0.23m, the charge 'q' on proton = 1.9 × 10^{-19} C and mass of proton = 1.67 × 10^{-27}Kg.

Thus,

  B = (1.67 × 10^{-27} × 3.5 ×10^{6}) ÷ (1.6 × 10^{-19} ×  0.23)

      = \frac{5.845 * 10^{-21} }{3.68 * 10^{-20} }

      = 0.15883

 B = 0.16 Tesla

The magnitude of the magnetic field 'B' is 0.16 Tesla.

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luda_lava [24]
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
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5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
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4 0
3 years ago
A negative charge is moved from point A to point B along an equipotential surface. Which of the following statements must be tru
elena-s [515]

Answer:

C) No work is required to move the negative charge from point A to point B.

Explanation:

An equipotential surface is defined as a surface connecting all the points at the same potential.

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The work done when moving a charge is given by

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where

q is the charge

\Delta V is the potential difference between the initial and final point of motion of the charge

However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so

\Delta V=0

And so, the work done is also zero.

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Answer:

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Explanation:

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

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5 0
2 years ago
Read 2 more answers
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

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