ANOTHER RUNNING DOG
Explanation:
In the given question it is to find a suitable reference point to describe the motion of dog. Here I could suggest that it is better to compare the dog with another running dog to create the relative speed difference to get a reliable motion variation.
Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to another dog which is already in motion.
Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with another dog running.
Think of it this way:
-- Any time you have something that means (some number) PER UNIT,
it doesn't matter how many units there are on the table or in the bucket,
because that amount doesn't change the (number) PER UNIT.
-- If oranges cost $1 PER POUND, it doesn't matter how many pounds
you buy, the whole bagful is still $1 PER POUND.
-- If a certain salad dressing has 40 calories PER Tablespoon, it doesn't
matter whether you eat a drop of it or drink the whole jar. You still get
40 calories PER Tablespoon.
-- Density means '(mass) PER unit of volume'. Whether you have a tiny
chip of the substance or a whole truckload of it, there's still the same
amount of mass IN EACH unit of volume.
Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;
Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Answer:
2.145×10^-10 V or 0.2145nV
Explanation:
From hf=eV
h= Plank's constant = 6.6×10^-34JS
f= frequency of the electromagnetic wave = 5.2×10^4 Hz
e= electronic charge= 1.6×10^-19 C
V= voltage
V= hf/e
V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C
V= 2.145×10^-10 V or 0.2145nV
Therefore the voltage created is 2.145×10^-10 V or 0.2145nV
My Physics teacher keeps giving us way too much work to do at home.
