Inertie is the resistance to change in motion so inertia depends solely on what?

In a thunderstorm, charge builds up on the water droplets or ice crystals in a cloud. thus, the charge can be considered to be distributed uniformly throughout the cloud. for the purposes of this problem, take the cloud to be a sphere of diameter 1.00 kilometer. the point of this problem is to estimate the maximum amount of charge that this cloud can contain, assuming that the charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air breaks down. this breakdown means that the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. the ionized air will then emit light due to the recombination of the electrons and atoms to form excited molecules that radiate light. in addition, the large current will heat up the air, resulting in its rapid expansion. these two phenomena account for the appearance of lightning and the sound of thunder. take the breakdown electric field of air to be eb=3.00ă—106n/c. part a estimate the total charge q on the cloud when the breakdown of the surrounding air is reached. express your answer numerically, to three significant figures, using ďµ0=8.85ă—10â’12c2/(nâ‹…m2) .

Astone is dropped from a cliff and falls 9.44 meters. what is the speed of the stone when it reaches the ground? a. 13.6 m/sec b. 1.39 m/sec c. 185 m/sec d. 9.80 m/sec

Acommon technique in analysis of scientific data is normalization. the purpose of normalizing data is to eliminate irrelevant constants that can obscure the salient features of the data. the goal of this experiment is to test the hypothesis that the flux of light decreases as the square of the distance from the source. in this case, the absolute value of the voltage measured by the photometer is irrelevant; only the relative value conveys useful information. suppose that in part 2.2.2 of the experiment, students obtain a signal value of 162 mv at a distance of 4 cm and a value of 86 mv at a distance of 5.7 cm. normalize the students' data to the value obtained at 4 cm. (divide the signal value by 162.) then calculate the theoretically expected (normalized) value at 5.7 cm.