3 - b) weight decreases
4 -a) 50 kg as mass is same everywhere
5 - b) 200 dynes
Given:
The force of attraction is F = 48.1 N
The separation between the charges is

Also, the magnitude of charge q1 = q2 = q.
To find the magnitude of charge.
Explanation:
The magnitude of charge can be calculated by the formula

Here, k is the Coulomb's constant whose value is

On substituting the values, the magnitude of charge will be

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.
Answer:
The answer to your question should be D.
Explanation:
reactants are on the laft side of arrow and products are on right side of arrow
Answer:
The answer is not B or D because they need light co2 and water to make their food so B and D are out. and unless someone is walking during winter to feed it. It will not be A so the answer is C.
Explanation:
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q = 
=
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is
.