Gravitational potential energy=mass*gravitational acceleration*height
Kinetic energy = 0.5*mass*velocity²
So with the given data
K.E 0.5*1*x²=12.5 v²=12.5÷(0.5*1)
v=√12.5÷(0.5*1) v=5
GPE mass*gravitational acceleration*height
1*9.81*h=98
h=98÷(9.81*1)
h= 9.98 m
Answer:
f = 1 / [ 2*pi*R*C*Sqrt(2*N)
Explanation:
f = 1 / [ 2*pi*R*C*Sqrt(2*N) is the formula. Sorry I answered in English.
Answer:
120,000J
Corrected question;
In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?
Explanation:
Given;
Total energy Et = 500,000 J
Wasted Energy Ew = 380,000J
The amount useful energy is the amount of energy that is available for supply.
This can be derived by subtracting the wasted energy from the total energy.
Useful energy = Total Energy - wasted energy
Eu = Et - Ew
Substituting the given values;
Eu = 500,000J - 380,000
Eu = 120,000 J
The amount of useful energy produced in one hour is 120,000 J
Was there any choice of answers so i can help or something u have to figure out
The electric force between the two particles are calculated through the equation,
F = kQ₁Q₂ / d²
where F is the force, k is a constant called Coulomb's law constant, Q₁ and Q₂ are the charges, and d is the distance. This equation is called the Coulomb's law.
It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.
The answer to this item is therefore letter A.