Answer:
75
Explanation:
Power is current times voltage:
P = IV
Voltage is current times resistance:
V = IR
Therefore:
P = I²R
Given I = 0.62 A and R = 195 Ω:
P = (0.62 A)² (195 Ω)
P ≈ 75 W
Answer:
We know that the force pulling the box in the positive x direction has a magnitude of m g sin 30 . Using Newtons Second Law, F = ma , we just need to solve for a :
ma=mgsin30
a=gsin30
=(10m/s2)(0.500)
=5m/s2
Um, this doesn't make any sense. By climbing a hill, you are decreasing your momentum and kinetic energy, so it slows you down. The only positive, is after you have climbed the hill, you have more potential energy, and it will be released once you go down the hill, but you will not be as fast as if you ignored the hill.
The speed of sound is greater in ice (4000 m/s), then in water (1500 m/s), then in air (340 m/s). The explanation for this is the differente state of the matter in the three cases.
In fact, sound waves travel faster in solids (like ice), then in liquids (like water), then in gases (like air). This is because the speed of the sound wave depends on the density of the medium: the greater the density, the faster the sound wave. This can be easily understood by thinking at how a sound wave propagates: a sound wave is a vibration of molecules, which is transmitted throughout the medium by collision of the molecules. Therefore, the smaller the spacing between the molecules (such as in solids), the more efficient is the propagation, and so the sound wave is faster. On the contrary, there is a large spacing between molecules in gases (such as in the air), so there are less collisions between the molecules and so the wave is not transmitted efficiently, and so it has less velocity.
Answer & Explanation:
Crashing into the asteroid would cause the satellite to slow down, stop, or reverse direction, because it is a force in the opposite direction to the satellite's motion. Whichever crash was a stronger force would cause it to change motion more. It takes a stronger force to change the velocity of a more massive object.