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Vikki [24]
3 years ago
14

Two tiny beads are 25 cm apart with no other charged objects or fields present. Bead A has a net charge of magnitude 10 nC and b

ead B has a net charge of magnitude 1 nC. Which one of the following statements is true about the magnitudes of the electric forces on these beads?A. The force on A is 100 times the force on B.B. The force on B is 10 times the force on A.C. The force on B is 100 times the force on A.D. The force on A is 10 times the force on B.E. The force on A is exactly equal to the force on B.
Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

E. The force on A is exactly equal to the force on B.

Explanation:

The force between two charges is given by

F=\dfrac{kq_1q_2}{r^2}

where

q_1 = Charge on particle 1

q_2 = Charge on particle 2

r = Distance between the charges

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

F=\dfrac{8.99\times 10^9\times 10\times 10^{-9}\times 1\times 10^{-9}}{(25\times 10^{-2})^2}\\\Rightarrow F=0.0000014384\ N

This force will be exerted on both the charges equally.

So, The force on A is exactly equal to the force on B

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Answer: D

Explanation: Force can do everything else.

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Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 time
skad [1K]

Answer:

g' = 13.5 m/s²

Explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re²   --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'²   --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

<u>g' = 13.5 m/s²</u>

6 0
3 years ago
Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,
Nitella [24]

<u>Answer:</u> The amount of heat needed is 29000 Cal.

<u>Explanation:</u>

The process involved in this problem are:

(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)

Now, we calculate the amount of heat released or absorbed in all the processes.

  • <u>For process 1:</u>

q_1=mC_{p,s}\times (T_2-T_1)

where,

q_1 = amount of heat absorbed = ?

m = mass of ice = 200 g

T_2 = final temperature = 0^oC

T_1 = initial temperature = -30^oC

Putting all the values in above equation, we get:

q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal

  • <u>For process 2:</u>

q_2=m\times L_f

where,

q_1 = amount of heat absorbed = ?

m = mass of water or ice = 200 g

L_f = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

q_2=200g\times 80Cal/g=16000Cal

  • <u>For process 3:</u>

q_3=m\times C_{p,l}\times (T_{2}-T_{1})

where,

q_3 = amount of heat absorbed = ?

m = mass of water = 200 g

T_2 = final temperature = 50^oC

T_1 = initial temperature = 0^oC

Putting all the values in above equation, we get:

q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal

Calculating the total heat absorbed, we get:

Q=q_1+q_2+q_3

Q=3000+16000+10000=29000Cal

Hence, the amount of heat needed is 29000 Cal.

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as potential energy = mgh .. here h becomes 0 so potential energy is 0 joules..</em>
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