Answer:
The angular magnification is 
Explanation:
From the question we are told
The focal length is 
The near point is 
The angular magnification is mathematically represented as

Substituting values

Answer:
option C
Explanation:
given,
energy dissipated by the system to the surrounding = 12 J
Work done on the system = 28 J
change in internal energy of the system
Δ U = Q - W
system losses energy = - 12 J
work done = -28 J
Δ U = Q - W
Δ U = -12 -(-28)
Δ U = 16 J
hence, the correct answer is option C
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Answer:
The speed is
.
(a) is correct option.
Explanation:
Given that,
Potential difference 
Speed 
If it were accelerated instead
Potential difference 
We need to calculate the speed
Using formula of initial work done on proton

We know that,


Put the value into the formula

....(I)
If it were accelerated instead through a potential difference of
, then it would gain a speed will be given as :
Using an above formula,

Put the value of 



Hence, The speed is
.
B is the answer, I’m really good at this subject