Answer:
Option A, B, C and D
Explanation:
First to all, we need to remember something. Mercury is the first planet to our solar system, therefore, it's the closest planet to the sun. Because of this, temperatures of that planet are way too high.
Mercury has a very thin atmosphere so it barely exists. It also has a low gravity and receives large gusts of solar winds from the Sun, that's why it has high temperature, and therefore, it's escape velocity is very low too.
Of course, it's one of the smallest planets in our solar system, so,the atmosphere of Mercury is unstable and constantly shifting. As the atmosphere’s materials are being made, they are also being taken away at the uppermost layers due to solar winds. The composition of the atmosphere can also change as you move across the planet.
Answer:
a)ΔS₁ = - 9.9 J/K
ΔS₂ = 69 J/K
b)The entropy change for the rod = 0 J/K
c)ΔS = 59.1 J/K
Explanation:
Given that
T₁ = 699 K
T₂= 101 K
Q= 6970 J
Change in entropy given as

For 699 K:


ΔS₁ = - 9.9 J/K ( Negative because heat is leaving from the system)
For 101 K;


ΔS₂ = 69 J/K
The entropy change for the rod = 0 J/K
Entropy change for the system
ΔS = ΔS₂ + ΔS₁
ΔS = 69 -9.9 J/K
ΔS = 59.1 J/K
Answer:
The time interval is 
Explanation:
From the question we are told that
The length of the string is 
The mass of the string is 
The tension on the string is 
The velocity of the pulse is mathematically represented as

Where
is the linear density which is mathematically evaluated as

substituting values


Thus


The time taken is evaluated as

substituting values


The correct option is A.
J.J Thompson proposed the plum model of atomic structure. The model states that the atom is a sphere of positively charged matter in which electrons are embedded.
Rutherford designed an experiment to verify the truth of this model. He set up a beam of alpha particles which are to be passed a thin gold foil. If the JJ model of atomic structure is correct, all of the alpha particles suppose to pass through the foil, but some of the alpha particles bounced back while those that passed through the foil emerged at different angles.
To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration

Here
= Angular acceleration
Initial and final angular velocity
There is not initial angular velocity,then

We know that the relation between the tangential velocity with the angular velocity is given by,

Here,
r = Radius
= Angular velocity,
Rearranging to find the angular velocity

Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,


Therefore teh average angular acceleration of each wheel is 