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3 years ago

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Need help!!! A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force a

cting on the charge?
A. 0.27 N
B. 2.7 N
C. 27 N
D. 270 N

2 answers:

Papessa [141]3 years ago

4 0

Answer:

Electric force, F = 2.7 N

Explanation:

It is given that,

Magnitude of electric charge, $q=8.5\times 10^{-6}\ C$

Electric field, $E=3.2\times 10^{5}\ N/C$

We have to find the electric force acting on the charge. It is given by the product of electric charge and electric field i.e.

$F=q\times E$

$F=8.5\times 10^{-6}\times 3.2\times 10^{5}$

F = 2.72 N

Hence, the correct option is (b) " 2.7 N"

ddd [48]3 years ago

3 0

B is the right answer. Multiply numbers you get the answer

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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

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Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

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Answer:

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Explanation:

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r = Radius of the equator

t = Time taken to complete one rotation = 24 h

Perimeter of a circle is given by

$P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi$

Angular speed is given by

$\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}$

Velocity if given by

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3 years ago

A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis

olga55 [171]

Answer:

(a) 0.942 m

(b) 18.84 m/s

(c) 2366.3 m/s²

(d) 0.05 s

Explanation:

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(b) Its frequency, f, is 1200 rev/min = $\dfrac{1200}{60}$ rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

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(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

$T = \dfrac{1}{f}$

T = 1/20 = 0.05 s

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