I am 11 year old and u just needed to login in so I can't answer ur questions
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
Answer:
material work function is 0.956 eV
Explanation:
given data
red wavelength 651 nm
green wavelength 521 nm
photo electrons = 1.50 × maximum kinetic energy
to find out
material work function
solution
we know by Einstein photo electric equation that is
for red light
h ( c / λr ) = Ф + kinetic energy
for green light
h ( c / λg ) = Ф + 1.50 × kinetic energy
now from both equation put kinetic energy from red to green
h ( c / λg ) = Ф + 1.50 × (h ( c / λr ) - Ф)
Ф =( hc / 0.50) × ( 1.50/ λr - 1/ λg)
put all value
Ф =( 6.63 × (3 × ) / 0.50) × ( 1.50/ λr - 1/ λg)
Ф =( 6.63 × (3 × ) / 0.50 ) × ( 1.50/ 651× - 1/ 521 ×)
Ф = 1.5305 × J × ( 1ev / 1.6 × J )
Ф = 0.956 eV
material work function is 0.956 eV
Answer:
air resistance, gravitational force
The tangent looks good.
The curve is a bit crooked, at the 0.9 and 1.
But overall, cool graph.