What uhhhhhhhhhhhhhhhhhhhhh
Answer:
v = 0.489 m/s
Explanation:
It is given that,
Mass of a box, m = 1.5 kg
The compression in the spring, x = 6.5 cm = 0.065 m
Let the spring constant of the spring is 85 N/m
We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.


So, the speed of the box is 0.489 m/s.
Answer:
208 Joules
Explanation:
The radius of the circular path the charge moves, r = 26 m
The magnetic force acting on the charge particle, F = 16 N
Centripetal force,
= m·v²/r
Kinetic energy, K.E. = (1/2)·m·v²
Where;
m = The mass of the charged particle
v = The velocity of the charged particle
r = The radius of the path of the charged particle
Whereby the magnetic force acting on the charge particle = The centripetal force, we have;
F =
= m·v²/r = 16 N
(1/2) × r ×
= (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.
∴ (1/2) × r ×
= (1/2) × 26 m × 16 N = = (1/2)·m·v² = K.E.
∴ 208 Joules = K.E.
The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.
Answer:
(a) Energy density will be equal to 
(b) Total energy will be equal to 0.0718 J
Explanation:
It is given that length of solenoid l = 78.8 cm = 0.788 m
Cross sectional area 
Number of turns of the wire N = 914
Current in the solenoid i = 8.25 A
Inductance of the wire is equal to 
(b) Total energy stored in magnetic field 
(a) Energy density will be equal to
