Complete Question
The complete question is is shown on the first uploaded
Answer:
The elastic potential energy at point B is
The kinetic energy at point D is
Explanation:
Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is
So at point B
KE at point B is 50J
So
Now at point D
at point D is 25J
So
Answer:
<h2>187,500N/m</h2>Explanation:
From the question, the kinectic energy of the train will be equal to the energy stored in the spring.
Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².
Equating both we will have;
1/2 mv² = 1/2ke²
mv² = ke²
m is the mass of the train
v is the velocity of then train
k is the spring constant
e is the extension caused by the spring.
Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m
Substituting this values into the formula will give;
30000*4² = k*1.6²
The value of the spring constant is 187,500N/m
Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by
The frequency of both speakers is
Generally the distance of the listener to the first speaker is mathematically represented as
Generally the distance of the listener to second speaker at its new position is
Generally the path difference between the speakers is mathematically represented as
Here is the wavelength which is mathematically represented as
=>
=>
=>
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=>
=>
Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
- L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb