Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.
Answer:
the final energy of the system is 35.5 kJ.
Explanation:
Given;
initial energy of the system, E₁ = 10 kJ
heat transferred to the system, q₁ 30 kJ
Heat lost to the surrounding, q₂ = 5kJ
heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ
work done on the system, W = 500 J = 0.5 kJ
Apply first law of thermodynamic,
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the heat gained by the system
W is work done on the system
ΔU = 25kJ + 0.5 kJ
ΔU = 25.5 kJ
The final energy of the system is calculated as;
E₂ = E₁ + ΔU
E₂ = 10 kJ + 25.5 kJ
E₂ = 35.5 kJ.
Therefore, the final energy of the system is 35.5 kJ.
Answer:
The answers to your questions are given below
Explanation:
22. The energy of an electromagnetic wave and it's frequency are related by the following equation:
E = hf
Where:
E => is the energy
h => is the Planck's constant
f => is the frequency
From the equation i.e E = hf, we can conclude that the energy of a wave is directly proportional to it's frequency. This implies that an increase in the frequency of the wave will lead to an increase in the energy of the wave and also, a decrease in the frequency will lead to a decrease in the energy of the wave.
23. Gamma ray and radio wave are both electromagnetic waves. All electromagnetic waves has a constant speed of 3×10⁸ m/s in space.
Thus, gamma ray and radio wave have the same speed in space.
I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
