Its a tightly-packed particles gain energy, allowing them to move more freely.
Alexander Calders mobiles, like untitled, move when air currents move through them, making them kinetic. Alexander Calder was an American artist, famous for his abstract sculpture, hanging mobiles, and Kinetic art. Kinetic art is the <span>art that depends on motion for its effect. The kinetic art's works of Alexander Calders were called "mobiles".</span>
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:
a. Wavelength = λ = 20 cm
b. Next distance of maximum intensity will be 40 cm
Explanation:
a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.
Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.
This we get the equation:
(nλ + λ/2) - nλ = 30-20
λ/2 = 10
λ = 20 cm
b. at what distance, sound intensity will be maximum again.
For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.
= 30 + λ/2
= 30 + 20/2
=30+10
=40 cm
<span>Raj is trying to make a diagram to show what he has learned about nuclear fusion.
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