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Doss [256]
2 years ago
11

What must the charge (sign and magnitude) of a particle of mass 1.40 gg be for it to remain stationary when placed in a downward

-directed electric field of magnitude 640 N/CN/C
Physics
1 answer:
NISA [10]2 years ago
4 0

Answer:

the charge of the particle is -2.144 x 10⁻⁵ C.

Explanation:

The force acting on the particle is calculated as;

F = EQ = mg

Q = \frac{mg}{E}

where;

Q is magnitude of the charge of the particle

Q = \frac{(1.4\times 10^{-3})(9.8)}{640} \\\\Q = 2.144 \ \times \ 10^{-5} \ C

since the magnetic field is acting downward, the force must be acting upward in opposite direction.

Thus, the charge of the particle will be -2.144 x 10⁻⁵ C.

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a snail can move approximately 0.30 meters per minute.How many meters can the snail cover in 15 minutes ?
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a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
3 years ago
A car travels a distance of 400 m in 5 seconds. Calculate its average velocity.
ikadub [295]

Answer:

80 m/s

Explanation:

x = 400 m

t = 5 s

x = vt,

400 = v(5),

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5 0
2 years ago
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The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr
Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

I(t)=0.88e^{-t/6\times3600s}

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

q=\int\limits {I} \, dt

Here the limits of integration is from 0 to infinite. So,

q=\int\limits {0.88e^{-t/6\times3600s}}\, dt

q=0.88\times(-6\times3600)(0-1)

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}

N = 1.2 x 10²³

8 0
3 years ago
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