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arlik [135]
3 years ago
13

A proton moves from left to right across the screen at speed v0. Then a uniform magnetic field of strength 0.80 T is directed in

to the screen. What is the direction of the force on the proton?

Physics
1 answer:
Anton [14]3 years ago
8 0

Answer: The force is  directed upward

Explanation: Considering the Lorentz force, given by:

F= qv×B

using the right hand rule and considering the direction of electron velocity and the magnetic field from the figure, the vectorial product gives a  force vector upwards .

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An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
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<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>

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<span>in the ft - lb - s system </span>

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3 years ago
Read 2 more answers
If transpiration didn't take place, would water be able to enter the roots of the plant? Explain your reasoning
s2008m [1.1K]

Answer:

If transpiration didn't take place water would still be able to enter the roots of a plant

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A gas is collected from a radioactive material; upon inspection, the gas is identified as helium. the presence of the helium ind
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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
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