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KATRIN_1 [288]
4 years ago
15

Two objects are electrically charged. The net charge on one object is doubled. Therefore, the electric force _____.

Physics
2 answers:
creativ13 [48]4 years ago
7 0
If net charge on one object is doubled, then electric force will also get double. It is because they are directly proportional to each other.

Hope this helps!
labwork [276]4 years ago
3 0

Answer:

The electric force gets doubled.

Explanation:

According to Coulomb's law the force acting on charged bodies is directly proportional to the product of charges and inversely proportional to the square of distance between them.

Mathematically, it can be written as :

F=k\dfrac{q_1q_2}{r^2}

Where

r is the distance between the charges

In this question, it is given that the net charge on one object is doubled. As we can see from Coulomb's law, the electric force (F') gets doubled.

i.e. F'=2\times k\dfrac{q_1q_2}{r^2}

F' =2 F

Hence, this is the required solution.

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C
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3 years ago
How much heat is required to raise 100 grams of water (c= 4.18) by 5 degrees Celsius?
Andrei [34K]

Answer:

Heat capacity, Q = 2090 Joules.

Explanation:

Given the following data;

Mass = 100 grams

Specific heat capacity = 4.18 J/g°C.

Temperature = 5°C

To find the quantity of heat required;

Heat capacity is given by the formula;

Q = mct

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

t represents the temperature of an object.

Substituting into the formula, we have;

Q = 100*4.18*5

Heat capacity, Q = 2090 Joules.

7 0
3 years ago
(a) calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the radius of the earth is
kakasveta [241]

(a) The value of the gravitational acceleration is given by:

g=\frac{GM}{r^2}

where

G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M is the Earth's mass

r=6371 km=6.371 \cdot 10^6 m is the Earth's radius at the pole

If we use the value for g given by the problem, g=9.830 m/s^2, and we rearrange the equation above, we find the value of the Earth's mass:

M=\frac{gr^2}{G}=\frac{(9.830 m/s^2)(6.371 \cdot 10^6 m)^2}{6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}}=5.982 \cdot 10^{24} kg

(b) The value we found for the Earth's mass is 5.982 \cdot 10^{24} kg, and we see that this value is slightly larger than the accepted value of 5.979 \cdot 10^{24} kg.

5 0
3 years ago
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