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Yuki888 [10]
3 years ago
11

Common static electricity involves charges ranging from nanocoulombs to microcoulombs. How many electrons must be removed from a

neutral object in order to leave a net charge of 9.7 μC?
Physics
1 answer:
madam [21]3 years ago
3 0

Answer:

6.0625 x 10^13

Explanation:

Charge of one electron, e = 1.6 x 10^-19 C

Total charge, Q = 9.7 μ C = 9.7 x 10^-6 C

The number of electrons is given by

n = Total charge / charge of one electron

n = Q / e

n = (9.7 x 10^-6) / (1.6 x 10^-19)

n = 6.0625 x 10^13

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If your parents were going through a divorce and you needed to talk to someone, who would be the best professional to see?Clinic
just olya [345]

psychologist counseling would be the correct answer I believe

3 0
3 years ago
Read 2 more answers
What is 0.0025 m^2 in cm^2? 1m = 100cm.
PSYCHO15rus [73]

Answer:

25 cm²

Explanation:

Meters and centimeters are both the units for measuring length. The SI unit of measuring length is meters.

Area is the quantity which measures the cross-section occupied by the object.

Thus,

Given that = Area = 0.0025 m²

To convert into cm²

1 m = 100 cm

So, 1 m² = 10000 cm²

So,

<u>Area = 0.0025 × 10000 cm² = 25 cm²</u>

6 0
3 years ago
A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of
miskamm [114]

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= \frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}

= 1.72 x 10³ N.

8 0
3 years ago
A resistor is connected to a 36v power supply. An ammeter measures a current of 2.0 A going through it. Determine the resistance
m_a_m_a [10]

R = 18 ohms

Explanation:

Given:

V = 36 volts

I = 2.0 A

R = ?

Use Ohm's law to solve for the resistance:

V = IR

or

R = V/I

= (36 volts)/(2.0 A)

= 18 ohms

8 0
2 years ago
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 10
Tema [17]

Answer:

Explanation:

(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.

Hence, his average velocity is 300m/150s=2ms^−1

(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.

Therefore, his average speed for this journey is 400m210s=1.9ms−1.

For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.

Hence, his average velocity for this case is 200m/210s=0.95ms^−1

7 0
3 years ago
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