The rotation of Earth is equivalent to one day which is comprised of 24 hours. To determine the number of miles in Earth's circumference, one simply have to multiply the given rate by the appropriate conversion factor and dimensional analysis. This is shown below.
C = (1038 mi/h)(24 h/1 day)
C = 24,912 miles
From the given choices, the nearest value would have to be 20,000 mile. The answer is the second choice.
It is know as smoke because if you cook food smoke will go up in the air and that makes vapor and also water from the ground it suck up
Answer:
An object which experiences either a change in the magnitude or the direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle at constant speed can be said to accelerate - the direction of the velocity changes.
if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.
Explanation:
The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.
Answer:
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
