Answer:
Explanation:
In case of gas , work done
W = ∫ p dV , p is pressure and dV is small change in volume
the limit of integration is from Vi to Vf .
= ∫ p dV
= ∫ p₀
dV
= p₀ 
/ ( 
)
= - 5p₀ 
Taking limit from Vi to Vf
W = - 5 p₀ ( 
) ltr- atm.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
Answer:
Net electric field, 
Explanation:
Given that,
Charge 1, 
Charge 2, 
distance, d = 3.2 cm = 0.032 m
Electric field due to charge 1 is given by :
Electric field due to charge 2 is given by :
The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :
So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.
A.) We use the famous equation proposed by Albert Einstein written below:
E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s
Substituting the value:
E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules
b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
