Answer:
Q = 29.4 x 10⁻⁹ C.
Explanation:
Electric field due to a charge Q at distance d is given by coulomb law as follows
Electric field
E = k Q /d²
where for air k which is a constant is 9 x 10⁹
Given E = 32.08 , d = 2.84 m
Putting these values in the relation above, we have
![[tex]32.08=\frac{9\times 10^9\times Q}{(2.84)^2}](https://tex.z-dn.net/?f=%5Btex%5D32.08%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%20Q%7D%7B%282.84%29%5E2%7D)
[/tex]
Q = 29.4 x 10⁻⁹ C.
Answer:
(a) Force must be grater than 283.87 N
(B) Force will be equal to 193.945 N
Explanation:
We have given mass of the crate m = 49.6 kg
Acceleration due to gravity 
Coefficient of static friction 
Coefficient of kinetic friction 
(a) Static friction force is given by 
So to just start the crate moving we have to apply more force than 283.87 N
(B) This force will be equal to kinetic friction force
We know that kinetic friction force is given by 
Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
The gravitational potential energy
gpe = mgh
By heat or change in matter,
there can be different reactions that create heat (like exothermic or endothermic reactions)
or movement can give off energy too,
energy is basically just heat
