Answer:
6.5e-4 m
Explanation:
We need to solve this question using law of conservation of energy
Energy at the bottom of the incline= energy at the point where the block will stop
Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=
Energy at the point where the block will stop consists of only gravitational potential energy=
Hence from Energy at the bottom of the incline= energy at the point where the block will stop
⇒
⇒
Also 
where
is the mass of block
is acceleration due to gravity=9.8 m/s
is the difference in height between two positions
⇒
Given m=2100kg
k=22N/cm=2200N/m
x=11cm=0.11 m
∴
⇒
⇒
⇒h=0.0006467m=
Answer:
The magnitude of the force, B = 5 Tesla, Up (North) direction
Explanation:
Magnetic force F= Eq where Electric field, E = 750 NC
and charge, q = -70 μC = -7 ×
C
F = 750 × -7 ×
F = 0.0525
But F = qvB; B = 
where B is the magnetic field
= 0.0525 ÷ ( -7 ×
× 30)
B = 5.0 Teslas
The force on a negative charge is in exactly the opposite direction to that on a positive charge.
Hence the direction of the charge is up (North).
Convection, because it is the process of heat transfer from one location to the next by the movement of fluids. The moving fluid carries energy within it.
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia, 
Initial angular velocity of the platform, 
Part A,
Let
is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

Here, 


Solving the above equation, we get the value as :

Part B,
The initial rotational kinetic energy is given by :



The final rotational kinetic energy is given by :



Hence, this is the required solution.
Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.