<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>
Answer:
t = 0.657 s
Explanation:
First, let's use the appropiate equations to solve this:
V = √T/u
This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.
Where:
V: Speed of the disturbance
T: Tension of the rope
u: linear density of the rope.
The density of the rope can be calculated using the following expression:
u = M/L
Where:
M: mass of the rope
L: Length of the rope.
We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:
u = 2.31 / 10.4 = 0.222 kg/m
Now, replacing in the first equation:
V = √55.7/0.222 = √250.9
V = 15.84 m/s
Finally the time can be calculated with the following expression:
V = L/t ----> t = L/V
Replacing:
t = 10.4 / 15.84
t = 0.657 s
"The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Column 2 would be on the y-axis."
typical plot is speed or velocity on the y-axis n time on the x-axis so the ans is Column 1 should be titled “Time,” and Column 2 should be titled “Velocity.”
We can calculate the density of the balloon as follows:
![\rho=\frac{mass}{volume}=\frac{1000g}{1000cm^3}=\frac{1g}{cm^3}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7Bmass%7D%7Bvolume%7D%3D%5Cfrac%7B1000g%7D%7B1000cm%5E3%7D%3D%5Cfrac%7B1g%7D%7Bcm%5E3%7D)
Therefore, the balloon will fall
Since the density of air is about 0.00123 g/cm^3 , the balloon is much more dense than the surrounding air. As a result, the balloon weighs more than the air that it displaces so the balloon will fall.
Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
Multiply each side
by 0.09 m³ : (4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg .
Force of gravity = (mass) x (acceleration of gravity)
= (360 kg) x (9.8 m/s²)
= (360 x 9.8) kg-m/s²
= 3,528 newtons .
That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).
Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.
The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.
So while it's in the water, the block seems to weigh
(3,528 - 882) = 2,646 newtons (about 595.2 pounds) .
But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.