Answer:
The temperature change of the copper is greater than the temperature change of the water.
Explanation:
deltaQ = mc(deltaT)
Where,
delta T = change in the temperature
m =mass
c = heat capacity

The temperature change in the copper is nearly 11 times the temperature change in the water.
So, the correct option is,
The temperature change of the copper is greater than the temperature change of the water.
Hope this helps!
Radio waves have many uses—the category is divided into many subcategories, including microwaves and electromagnetic waves used for AM and FM radio, cellular telephones and TV.
The lowest commonly encountered radio frequencies are produced by high-voltage AC power transmission lines at frequencies of 50 or 60 Hz. These extremely long wavelength electromagnetic waves (about 6000 km) are one means of energy loss in long-distance power transmission.
Extremely low frequency (ELF) radio waves of about 1 kHz are used to communicate with submerged submarines. The ability of radio waves to penetrate salt water is related to their wavelength (much like ultrasound penetrating tissue)—the longer the wavelength, the farther they penetrate. Since salt water is a good conductor, radio waves are strongly absorbed by it; very long wavelengths are needed to reach a submarine under the surface.
HOPE THIS REALLY HELPS YOU.
THANK YOU.
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is

.
In fact, the laws of motions on both x- and y- directions are


From the second equation, we get the time t at which the projectile hits the ground, by requiring

, and we get:

And inserting this value into Sx(t), we find

And this value is maximum when

, so this is the angle at which the projectile reaches its maximum distance.
So now we can take again the law of motion on the x-axis

And by using

, we find the value of the initial velocity v0:
Answer:
C. position
Explanation:
potential energy is based on stored energy, therefore it is C