Answer:0.153 Hz
Explanation: The relation between Time Period(T) and frequency(f) is given by T=1/f
Plug in the values and u arrive at the answer
Answer:
t = 0.657 s
Explanation:
First, let's use the appropiate equations to solve this:
V = √T/u
This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.
Where:
V: Speed of the disturbance
T: Tension of the rope
u: linear density of the rope.
The density of the rope can be calculated using the following expression:
u = M/L
Where:
M: mass of the rope
L: Length of the rope.
We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:
u = 2.31 / 10.4 = 0.222 kg/m
Now, replacing in the first equation:
V = √55.7/0.222 = √250.9
V = 15.84 m/s
Finally the time can be calculated with the following expression:
V = L/t ----> t = L/V
Replacing:
t = 10.4 / 15.84
t = 0.657 s
Answer:
Total work done in expansion will be 
Explanation:
We have given pressure P = 2.10 atm
We know that 1 atm 
So 2.10 atm 
Volume is increases from 3370 liter to 5.40 liter
So initial volume 
And final volume 
So change in volume 
For isobaric process work done is equal to 
So total work done in expansion will be
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Answer:
F. 25.82 s
Explanation:
Given:
Δy = 90 m
v₀ = 0 m/s
a = 0.27 m/s²
Find: t
Δy = v₀ t + ½ at²
90 m = (0 m/s) t + ½ (0.27 m/s²) t²
t = 25.82 s
