1. The velocity decreases, and the kinetic energy decreases.
2. An increase in temperature difference between the inside and outside of the building.
3. The total kinetic energy remains the same.
4. 76,761 J
5. The energy loss must increase.
Answer:
A) 138.8g
B)73.97 cm/s
Explanation:
K = 15.5 Kn/m
A = 7 cm
N = 37 oscillations
tn = 20 seconds
A) In harmonic motion, we know that;
ω² = k/m and m = k/ω²
Also, angular frequency (ω) = 2π/T
Now, T is the time it takes to complete one oscillation.
So from the question, we can calculate T as;
T = 22/37.
Thus ;
ω = 2π/(22/37) = 10.5672
So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g
B) In simple harmonic motion, velocity is given as;
v(t) = vmax Sin (ωt + Φ)
It is from the derivative of;
v(t) = -Aω Sin (ωt + Φ)
So comparing the two equations of v(t), we can see that ;
vmax = Aω
Vmax = 7 x 10.5672 = 73.97 cm/s
Consider a long train moving at speed v. Now consider a passenger throwing a ball inside this train, towards the back of the train, with same velocity v (but in the opposite direction of the train movement).
- A passenger inside the train will see the ball moving with speed v
- For an observer outside the train, however, the ball will appear as still. In fact, for him the ball will have a speed v (given by the movement of the train) -v (velocity of the ball but moving in the opposite direction), so the net velocity will be v+(-v)=0.
The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.
<h3>
UF cable</h3>
UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.
For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.
Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.
Learn more about UF cable here: brainly.com/question/8591560
