<u><em>Correct question:</em></u><u><em>a </em></u><u><em>200 g </em></u><u><em>ball is dropped from a height of 2m, bounces on a hard floor and rebounds to a height of 1.5m. What maximum force does the floor exert on the ball?</em></u>
<u><em>The diagram of the question is in the attachment.</em></u>
Answer:
Explanation:
V=√2gh
g-10m/s²
let u be initial velocity ad v be final velocity,
u=√2*10*2
u=6.324m/s
v=√2*10*1.5
v=5.477m/s
from the diagram
t=5ms=0.005s
F=-33.87N (the negative shows direction)
From the diagram Fmax=2F
Fmax= 2*33.87
=67.74N
Answer:
(d) 2.93 x 10⁻³ N
Explanation:
Given;
current in the wire, I = 11.4 A
angle of inclination, θ = 11.4⁰
magnetic field on the wire, B = 11. 4 x 10⁻³
length of the wire, L = 11.4 cm = 0.114 m
The magnitude of magnetic force on the wire is given by;
F = BILsinθ
F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)
F = 0.00293 N
F = 2.93 x 10⁻³ N
Therefore, the correct option is "D"
Answer:
F = 2325 N
Explanation:
given,
mass of the car, m = 1550 Kg
initial speed, u = 10 m/s
final speed, v = 25 m/s
time, t = 10 s
Average net force = ?
acceleration of the car
a = 1.5 m/s²
we know
F = m a
F = 1550 x 1.5
F = 2325 N
Net average force applied on the car is equal to 2325 N.
Answer:photon energy
Explanation:
Photon energy is energy carried by a single photon