I think so it ans wold be d.because if the source approaches the observer more and more wavefront will pass and it get squeezed.so wavelength decreases and frequency increases.
Answer:
It traveled 4 centimeters.
Explanation:
In a speed versus time graph, the distance travelled is given by the area under the graph.
In this graph we have the following:
- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s
- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s
Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:
Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Answer: I am pretty sure that you should pick radio waves.
Explanation: The scientist should use radio waves. I think this because you can use the radio waves to analyze the signals from outer space. This will work much better than anything there, to analyze it the best possible.
The best I could do.
