Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:
sin 65 = h / 150
Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
Explanation:
Here's a clearer rendering of the question requirements;
Complete the sentences with the correct wording. Imagine that a force gauge is mounted between the rope and the chain carousel saddle. If you do not touch your feet to the ground when the vehicle is stationary, the dynamometer indicates A / B. When the carousel turns, you will read C / D on the dynamometer.
A. Your weight with the saddle
C. Rope strength value
B. Your weight
D. Centripetal force value
Answer:
The distance between the earth and the star is increasing.
Explanation:
When we observe an object and its electromagnetic radiation has been displaced to blue, it means that it is getting closer to us, causing the light waves it emits to get closer together and its wavelength to decrease towards blue, this is knowm as blueshift.
On the contrary, when an object is rapidly moving away from us, the light waves or electromagnetic radiation it emits have been stretched from their normal wavelength to a longer wavelength, towards the red part of the spectrum. This is known as redshift.
This phenomenon of changes in wavelength and frequency due to movement (whether the source approaches or moves away) is described by the Doppler effect.
So for this case because the light we perceive from the star has moved to the red part of the visible spectrum, we can conclude that it is moving away from the earth, and that the distance between the star and the earth is increasing.
Answer:
Ecu/Eag = 0.46
Explanation:
E = PI/A
Ecu = Pcu × I/A
Pcu = 1.72×10^-8 ohm-meter
r = 0.8 mm = 0.8/1000 = 8×10^-4 m
A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π
Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1
Eag = Pag × I/A
Pag = 1.47×10^-8 ohm-meter
r = 0.5 mm = 0.5/1000 = 5×10^-4 m
A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π
Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π
Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46
