<span>Let's first off calculate the kinetic energy using the formula 1/2MV^2. Where the mass, M, is 0.6Kg. And speed, V, is 2. Hence we have 1/2 * 0.6 * 2^2 = 1.2J. Since kinetic energy is energy due to motion; hence at point B the rubber has a KE of 1.2J and not 7.5J. So I would say that only the Mass and speed is actually true; While it's kinetic energy is not true.</span>
Answer:
The mass of the object on the Moon (and anywhere else) is about 30.61kg. Please see more detail below.
Explanation:
Weight is the gravitational force exerted on the object and is a function of mass and gravitational acceleration:
(weight) = (mass) x (gravitational acceleration)
We are to find the mass, knowing the weight on Earth to be 300N:
(mass) = (weight on Earth) / (gravitational acceleration on Earth) = 300N / 9.8 m/s^2 = 30.61 kg
The mass of the object is 30.61kg.
The mass of the object is independent of gravity. Therefore the answer to the question "What is its mass on the Moon" is 30.61kg.
If the question were what is its weight on the Moon, the answer would be
(weight on Moon) = (mass) x (grav.accel. on Moon) = 30.61kg x 1.62 m/s^2 = 49.59N
which is about 1/6 of the object's weight on the Earth.
Answer:
8.9
Explanation:
We can start by calculating the initial elastic potential energy of the spring. This is given by:
(1)
where
k = 35.0 N/m is the initial spring constant
x = 0.375 m is the compression of the spring
Solving the equation,
Later, the professor told the student that he needs an elastic potential energy of
U' = 22.0 J
to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:
Therefore, Tom needs to increase the spring constant by a factor:
We know the formula for density = Mass/ volume
So Mass, M = Volume * Density
Volume = 3.5 L= 0.0035
Density = 1.50 g/ml = 1500 
Mass, M = 0.0035*1500 = 5.25 kg
So mass of liquid having density 1.50 g/ml and volume 3.5 liters is 5.25 kg.
Explanation:
Gauss Law relates the distribution of electric charge to the resulting electric field.
Applying Gauss's Law,
EA = Q / ε₀
Where:
E is the magnitude of the electric field,
A is the cross-sectional area of the conducting sphere,
Q is the positive charge
ε₀ is the permittivity
We be considering cases for the specified regions.
<u>Case 1</u>: When r < R
The electric field is zero, since the enclosed charge is equal to zero
E(r) = 0
<u>Case 2</u>: When R < r < 2R
The enclosed charge equals to Q, then the electric field equals;
E(4πr²) = Q / ε₀
E = Q / 4πε₀r²
E = KQ /r²
Constant K = 1 / 4πε₀ = 9.0 × 10⁹ Nm²/C²
<u>Case 3</u>: When r > 2R
The enclosed charge equals to Q, then the electric field equals;
E(4πr²) = 2Q / ε₀
E = 2Q / 4πε₀r²
E = 2KQ /r²
