You would want to include qualitative and quantitative evidence
To solve this problem we will apply the concepts related to energy conservation. For this purpose we will have that all the changes occurred in the energy change will be equivalent to the change in the potential and kinematic energies of the body. At the same time we will consider that the change in the final energy of the system will be reflected in the work of the system, therefore,


Here,
F = Force
m = mass
v = Velocity
h = Height
d = Distance



Replacing we have,


Therefore the launch energy is 365061ft-lb
Quantity of charge unloaded = 51.9 Coulombs
Potential difference = 10.93 megavolts = 1.093 x 10⁷ volts
1 volt = 1 joule per coulomb
1.093 x 10⁷ volts = 1.093 x 10⁷ joules per coulomb
Energy = (1.093 x 10⁷ J/C) x (51.9 C)
Energy = 567.267 megaJoules
That's <em>5.67 x 10⁸ Joules</em> .
==> My wife's blow-dryer is marked 1260 watts.
If the energy in this lightning strike could be collected, bottled, stored, and used as needed, it could run my wife's blow-dryer for 125 hours. That would save us more than $30 on our electric bill !
Answer:
Option (b) is correct.
Explanation:
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C, is called specific heat of the substance.
The formula of the specific heat is
H = m c (T' - T)
where, m is the mass, c is the specific heat and T' - T is the change in temperature.
So, to know the rise in temperature, by adding the known amount of heat, the specific heat is required.
So, option (b) is correct.
Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:

where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,

Now, we will calculate the value of g on the neutron star:

where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,

Therefore, the weight on the surface of the neutron star will be:

<u>W' = 1.66 x 10¹⁴ N</u>