Potential energy<span> is the </span>energy<span> that is stored in an object due to its position relative to some zero position. It is calculated by the expression PE = mgh where mg is the weight of the book and h is the height. It is calculated as follows:
PE = 50(1) = 50 J
</span>PE = 50(1.5) = 75 J
PE = 50(2) = 100 J
Answer:
The magnitude of the applied torque is
(e) is correct option.
Explanation:
Given that,
Mass of object = 3 kg
Radius of gyration = 0.2 m
Angular acceleration = 0.5 rad/s²
We need to calculate the applied torque
Using formula of torque
Here, I = mk²
Put the value into the formula
Hence, The magnitude of the applied torque is
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
Answer:
-2.3 × 10^-9 Coulombs(C).
Explanation:
So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;
=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "
=> " Assuming the shuttle is a conducting sphere of radius 15 m".
So, in order to estimate the value for the charge we will be making use of the equation below:
Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.
Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.
So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.
= -2.3 × 10^-9 Coulombs(C).
Answer:
v= 0.2 m/s
Explanation:
Given that
m₁ = 50 kg
m₂ = 100 g = 0.1 kg
u =10 0 m/s
If there is no any external force on the system then the total linear momentum of the system will be conserve.
Initial linear momentum = Final momentum
m₁u₁ + m₂ u₂ =m₂ v₂ +m₁v₁
50 x 0 + 0.1 x 100 = 50 v + 0
0+ 10 = 50 v
v= 0.2 m/s
Therefore the recoil speed will be 0.2 m/s.