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wlad13 [49]
3 years ago
11

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T. What i

s the magnitude of the magnetic force acting on the charge? 1.6 × 10–1 N 6.0 × 10–1 N 1.6 × 105 N 6.0 × 105 N
(please show your work in solving this)
Physics
2 answers:
Tamiku [17]3 years ago
6 0

Force on a moving charge is given by

F = q(v x B)

Given that

q = 6.8 C

v = 6.5 * 10^4 m/s

B = 1.4 T

angle = 15 degree

F = qvB sin\theta

F = 6.8*6.5 * 10^4 * 1.4 * sin15

F = 1.6 * 10^5 N

Mrac [35]3 years ago
6 0

Answer:

A

Explanation:

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Two conducting spheres of different sizes are at the same potential. The radius of the larger sphere is four times larger than t
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Answer:

0.8

Explanation:

The two spheres have the same potential, V.

Let the radius of the larger sphere be R and the radius of the smaller sphere be r,

=> R = 4r

Let the charge on the smaller sphere be q. Hence, the larger sphere will have charge Q - q.

The potential of the smaller sphere will be:

V_S = \frac{kq}{r}

The potential of the larger sphere will be:

V_L = \frac{k(Q - q)}{R}

Inputting R = 4r,

V_L = \frac{k(Q - q)}{4r}

Since V_S = V_L = V,

\frac{k(Q - q)}{4r} = \frac{kq}{r}

=> Q - q = 4q

=> 5q = Q

q = 0.2Q

The fraction of the charge Q that rests on the smaller sphere is 0.2

The charge of the larger sphere is:

Q - q = Q - 0.2Q = 0.8Q

∴ The fraction of the total charge Q that rests on the larger sphere is 0.8

7 0
3 years ago
Water enters a shower head
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Answer:

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Why it is advise to clean the ends of connecting wires before connecting them?
Svetradugi [14.3K]
If dirt and grease were good conductors of electrical current, then we could make wire
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3 years ago
Read 2 more answers
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular
RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

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angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0
3 years ago
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