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
Below are the choices that can be found from other source:
a. frequency
b. the Doppler effect
c. the resonance effect
<span>d. intensity
The answer is B.
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Yes
Explanation:
It is a reasonable result obtained.
Error = true value - measured value
true value = 24.5
measured value = 24.2
Error = 24.5 - 24.2 = 0.3g
The error reported in the reading is 0.3g
The reason why we had a disparity in the figures obtained from this measurement is primarily due to some erroneous scale.
The mixture at the end of the day is a solution.
We are expected to have the same mass but due to experimental or some form of random error introduced, we noticed a difference.
The value obtained is quite logical as we only had a deviation of 0.3g.
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Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
<u>f₂ = 3 Hz</u>
Answer: A pendulum is an object hung from a fixed point that swings back and forth under the action of gravity.
Example: playground swing
