Light from the stars, because the orbits make it difficult to see them.
Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
![E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%3D%20gravity%5Bm%2Fs%5E%7B2%7D%20%5D%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cm%3D%20%5Cfrac%7BE_%7Bp%7D%20%7D%7Bg%2Ah%7D%5C%5C%20m%3D%20%5Cfrac%7B100%7D%7B9.81%2A10%7D%5C%5C%5C%5Cm%3D%201.01%5Bkg%5D%5C%5C%5C%5C)
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
![P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\](https://tex.z-dn.net/?f=P_%7Be%7D%20%3Dm%2Ag%2Ah%5C%5CP_%7Be%7D%20%3D1.01%2A9.81%2A5%5C%5C%5C%5CP_%7Be%7D%20%3D%2050%20%5BJ%5D%5C%5C)
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]
Let us say that x is the cut that we will make on the
sides to make a box, therefore the new dimensions are:
l = 15 – 2x
w = 8 – 2x
It is 2x since we cut on two sides.
We know that volume is:
V = l w x
V = (15 – 2x) (8 – 2x) x
V = 120x – 30x^2 – 16x^2 + 4x^3
V = 120x – 46x^2 + 4x^3
Taking the 1st derivative:
dV/dx = 120 – 92x + 12x^2
Set dV/dx = 0 to get maxima:
120 – 92x + 12x^2 = 0
Divide by 12:
x^2 – (92/12)x + 10 = 0
(x – (92/24))^2 = -10 + (92/24)^2
x - 92/24 = ±2.17
x = 1.66, 6
We cannot have x = 6 because that will make our w
negative, so:
x = 1.66 inches
So the largest volume is:
V = 120x – 46x^2 + 4x^3
V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3
V = 90.74 cubic inches
Answer:

Explanation:
Given that,
Heat required, Q = 1200 J
Mass of the object, m = 20 kg
The increase in temperature, 
We need to find the specific heat of the object. The heat required to raise the temperature is given by :

So, the specific heat of the object is
.
Answer:
563.86 N
Explanation:
We know the buoyant force F = weight of air displaced by the balloon.
F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m
So, F = ρgV = ρg4πr³/3
substituting the values of the variables into the equation, we have
F = 1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3
= 1691.58 N/3
= 563.86 N