Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t
1 answer:
Answer:
8F_i = 3F_f
Explanation:
When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.
Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.
Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.
The electrostatic force, Fi, in the initial configuration can be calculated as follows.
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Answer:
Speed of the spacecraft right before the collision: .
Assumption: the earth is exactly spherical with a uniform density.
Explanation:
This question could be solved using the conservation of energy.
The mechanical energy of this spacecraft is the sum of:
- the kinetic energy of this spacecraft, and
- the (gravitational) potential energy of this spacecraft.
Let denote the mass of this spacecraft. At a distance of
from the center of the earth (with mass
), the gravitational potential energy (
) of this spacecraft would be:
.
Initially, (the denominator of this fraction) is infinitely large. Therefore, the initial value of
will be infinitely close to zero.
On the other hand, the question states that the initial kinetic energy () of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.
Right before the collision, the spacecraft would be very close to the surface of the earth. The distance between the spacecraft and the center of the earth would be approximately equal to
, the radius of the earth.
The of the spacecraft at that moment would be:
.
Subtract this value from zero to find the loss in the of this spacecraft:
Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the of this spacecraft would be equal to the size of the gain in its
.
Therefore, right before collision, the of this spacecraft would be:
.
On the other hand, let denote the speed of this spacecraft. The following equation that relates
and
to
:
.
Rearrange this equation to find an equation for :
.
It is already found that right before the collision, . Make use of this equation to find
at that moment:
.
Answer:
C) Pressure will compress a gas, reducing its volume and giving it a greater density and concentration of particles.
Explanation:
At constant temperature, pressure and volume are inversely related.
P V = constant
As the pressure increases, the gas compresses, the particles come closer reducing the volume of gas.
As we know, with decrease in volume, density increases.
Thus, the pressure of a gas is directly related to concentration of particles. Increase in pressure causes increase in concentration of the particles.
Answer:
The wavelength of the visible line in the hydrogen spectrum is 434 nm.
Explanation:
It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.
For Balmer series, the wave number is given by :
R is the Rydberg's constant
For Balmer series, n₁ = 2. So,
or
So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.