Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
=
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
Answer:
D
Explanation:
She says that the object of the experiment is to see how far the string stretches given a mass attached to the string.
The only thing that is at issue is either the mass or the amount the string stretches.
Nothing else matters.
The dependent variable therefore is the amount the string stretches. So the last choice is the answer.
If you illustrate the problem, you will somewhat come up with the figure shown. The missing value is the hypotenuse of the right triangle. Using the pythagorean theorems, the value is determined to be
x = √(0.7^2 + 2.4^2)
x = 2.5 km
2.5 km is the magnitude of the distance. If you want to incorporate the displacement, the answer is reported as
2.5 km, southeast. The direction is determined from the starting point to the endpoint.
Apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 180kPa, T = -8.0°C = 265.15K
Final P and T values:
P = 245kPa, T = ?
Set the initial and final P/T values equal to each other and solve for the final T:
180/265.15 = 245/T
T = 361K