The answer is d. Wavelength
Answer:
We kindly invite you to read carefully the explanation and check the image attached below.
Explanation:
According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:
(1)
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Initial time, measured in seconds.
- Final time, measured in seconds.
Now we obtain the kinematic equations for thrust and free fall stages:
Thrust (
,
,
,
)
(2)
Free fall (
,
,
,
)
(3)
Now we created the graph speed-time, which can be seen below.
Refer to the diagram shown below.
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N
Answer:
the smallest radius of the circular path is 8.1 km
Explanation:
The computation of the smallest radius of the circular path is given below:
Given that
V = Velocity = 201 m/s
a_c = acceleration = 5 m/s^2
radius = ?
As we know that
a_c = V^2 ÷ r
5 = 201^2 ÷ r
r = 201^2 ÷ 5
= 8,080.2 g
= 8.1 km
Hence, the smallest radius of the circular path is 8.1 km