Explanation:
The given data is as follows.
Mass of small bucket (m) = 4 kg
Mass of big bucket (M) = 12 kg
Initial velocity (
) = 0 m/s
Final velocity (
) = ?
Height
= 2 m
and,
= 0 m
Now, according to the law of conservation of energy
starting conditions = final conditions

235.44 =
+ 78.48
= 4.43 m/s
Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.
To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are
Comet mass 
Radius 
Rock was dropped from a height 'h' from surface = 1m
The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

Where G means gravitational universal constant and M the mass of the planet


Now calculate the value of the time




The time taken for the rock to reach the surface is t = 87.58s
Answer:
Speed of 1.83 m/s and 6.83 m/s
Explanation:
From the principle of conservation of momentum
where m is the mass,
is the initial speed before impact,
and
are velocity of the impacting object after collision and velocity after impact of the originally constant object
Therefore
After collision, kinetic energy doubles hence
Substituting 5 m/s for
then
Also, it’s known that
hence
Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then
Substituting,
Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
= rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV