Answer: Magnitude of the force exerted on the egg by the ground is 9.2N
Explanation:
Given the following :
Mass of egg (m) = 150g = 0.15kg
Height(h) from which egg is dropped = 3m
velocity of egg before hitting the ground (u) = 4.4m/s
Final velocity of egg (V) = 0
Time taken (t) = 0.072s
Magnitude of the force exerted on the egg by the ground can be found by applying Newton's 2nd law:
Momentum = mass × velocity
From Newton's second law:
Force = mass × change in Velocity with time ;
That is
F = m * ΔV / t
Inputting our values
F = 0.15 * (4.4 - 0) / 0.072
F = 0.15 × (4.4 / 0.072)
F = 0.15 × 61.11
F = 9.16N
F = 9.2N
Answer:
59.4 meters
Explanation:
The correct question statement is :
A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?
Solution:
We know for a circle of radius r and θ angle by an arc of length S at the center,
S=rθ
This gives
θ=S/r
also we know angular velocity
ω=θ/t where t is time
or
θ=ωt
and we know
1 revolution =2π radians
From this we have
angular velocity ω = 1.4 revolutions per sec = 1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec
Putting values of ω and time t in
θ=ωt
we have
θ= 8.8 rad / sec × 4.5 sec
θ= 396 radians
We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)
put this value of θ and r in
S=rθ
we have
S= 396 radians ×0.15 m=59.4 m
Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.
