Phases of the moon. Help!???????
1 answer:
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Answer:
a
The output power is
b
The Amplitude would decrease by
Explanation:
From the question we are told that
The diameter of the steel wire is = 1.0mm
The raduis of this steel wire is
Now from the question we can deduce that the power output is equal to the power being transmitted by wave on the wire this is mathematically represented as
Where is the mass per unit length of the wire
This is mathematically evaluated as
Where a is the area of the the wire =
is the density of steel with a generally value of
So
is velocity of the wave
This is mathematically evaluated as
substituting 60Hz for f
We have
is the amplitude with a given value of 0.50 cm
v is the linear velocity of the wave
This is mathematically evaluated as
Where T is the tension with a given value of
Substituting values into equation 1
Since the doubling of the frequency does not affect the amplitude and from equation one the output power is of the Amplitude, Then the Amplitude would decrease by
Answer:
b. less than w.
Explanation:
In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).
This is supported by the length contraction equation below:
L =
Here, L is the observed length
is the original length of the object
v is the relative speed between the object and the observer
and c is the speed of light
Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0
This effectively decreases the length that is observed.
Refer to the free body diagram shown melow.
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N