-- Before Adrian left the airplane, his gravitational potential energy was
(mass) x (gravity) x (height) = (80kg) x (9.81m/s²) x (1,000m) = 784,800 joules
-- When he reached the ground, his kinetic energy was
(1/2) x (mass) x (speed)² = (40kg) x (5m/s)² = 1,000 joules
-- Between the airplane and the ground, the Adrian lost
(784,800 joules) - (1,000 joules) = 783,800 joules
Where did all that energy go ?
Energy never just disappears. If it's missing, it had to go somewhere.
The Adrian used 783,800 joules of energy to push air our of his way
so that he could continue his parachute jump, and reach the ground
in time to be home for dinner.
The specific gravity is how the density of the object compares to the density of water. Water's density is 1gram per milliliter. We just need to figure out the density of the object.
The object is .8 kg and it displaces 500mL of water, so the density is the mass divided by the volume. Since the density of water is given in grams, we have to convert the objects mass from kg to g and then we can get the density.
.8kg * 1000g/kg = 800 grams
So
800g/500ml = 1.6grams/mL this is the density.
So divide the density of your object by the density of water, which is 1g/mL, you get 1.6 as the specific gravity. This means the object is 1.6 times more dense than water.
Answer:
Yes, it would work.
Explanation:
From Flehmings left hand rule, current can be generated when a coil cuts the magnetic field of a powerful magnet. Thus, the spin of properller turns a generator thereby converts motion to electrical energy.
The major challenge would be how to set the car in motion when at rest. But this can be solved by energy consrvation process. The law of conservation of energy states that energy cannot be created or destoyed, but transformed from one form to another. Thus, there would be mechanism having a device called an inverter which stores electric energy when the vehicle is in motion. This regenerates the required initial energy to set the electric car in motion when at rest or stops.
Answer:
a) 1.6*10^6 V
b) 13.35*10^6 V
Explanation:
The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:
(1)
q1 = 3.90µC = 3.90*10^-6 C
q2 = -2.4µC = -2.4*10^-6 C
r1 = 1.25 cm = 0.0125 m
r2 = -1.80 cm = -0.018 m
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
You replace all the parameters in the equation (1):
![V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V](https://tex.z-dn.net/?f=V%3Dk%5B%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%5D%5C%5C%5C%5CV%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0125m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.018m%7D%5D%3D1.6%2A10%5E6V)
hence, the total electric potential is approximately 1.6*10^6 V
b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:
r1 = 0.0150m - 0.0125m = 0.0025m
r2= 0.015m + 0.018m = 0.033m
Then, you replace in the equation (1):
![V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V](https://tex.z-dn.net/?f=V%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0025m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.033m%7D%5D%3D13.35%2A10%5E6V)
hence, for y = 1.50cm you obtain V = 13.35*10^6 V
Hi there!
Recall the equation for weight.
W = Weight (N)
M = Mass (kg)
g = acceleration due to gravity (m/s²)
The weight of an object depends upon its MASS and the strength of the GRAVITATIONAL field. We can solve for weight:
