Answer:
q = 0.0392 / V, for V= 0.1V q = 0.392 C
Explanation:
For this exercise we can assume that the power energy of the drops is transformed into kinetic energy, therefore we use the conservation of energy
starting point
Em₀ = U = q V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
q V = ½ m v²
q =
let's calculate
q = ½ 0.40 10⁻³ 14² / V
q = 0.0392 / V
The object to be painted is connected to ground therefore its potential is dro, but the gun where it is painted has a given potential, suppose it is
V = 0.1 V
q = 0.0392 / 0.1
q = 0.392 C
Kepler's third law of planetary motion states that:
"The ratio between the cube of the orbital radius of the planet and the square of the orbital period is constant". In formulas:
![\frac{r^3}{T^2} = k](https://tex.z-dn.net/?f=%20%5Cfrac%7Br%5E3%7D%7BT%5E2%7D%20%3D%20k)
where r is the orbital radius and T the orbital period.
Since this ratio is constant for every planet, we see that when the orbital radius r is larger (i.e. when the planet is farther from the Sun), the orbital period T is larger: this means the planet takes more time to complete one revolution around the Sun, so it moves slower.
Therefore, the correct option is:
<span>A planet moves slowest when it is farthest from the sun.</span>
'C' and 'D' are the same statement, and none of the 3 choices
is good.
Electrical energy is produced in a generator, a solar panel, or
a battery, not in things with resistance.
From the generator or battery, current flows through a circuit of
one or more components.
The greater the resistance of a component, the more energy is
LOST as the current flows through it. The component dissipates
the energy in the form of heat.
Answer:
The angle is ![141.058^{\circ}](https://tex.z-dn.net/?f=141.058%5E%7B%5Ccirc%7D)
Solution:
As per the question:
Mass of the objects are same, say m Kg each
Let their initial speed be 'u'
Both the objects move apart from each other at
m/s
Now, the angle between the objects' initial velocities is given by using the law of conservation of momentum:
Now, applying the principle of momentum conservation along the horizontal axis:
![mucos\theta + mucos\theta = 2m\frac{u}{3}](https://tex.z-dn.net/?f=mucos%5Ctheta%20%2B%20mucos%5Ctheta%20%3D%202m%5Cfrac%7Bu%7D%7B3%7D)
![cos\theta = \frac{1}{3}](https://tex.z-dn.net/?f=cos%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B3%7D)
![theta = cos^{- 1}(\frac{1}{3}) = 70.528^{\circ}](https://tex.z-dn.net/?f=theta%20%3D%20cos%5E%7B-%201%7D%28%5Cfrac%7B1%7D%7B3%7D%29%20%3D%2070.528%5E%7B%5Ccirc%7D)
Now, momentum conservation along vertical axis:
![musin\theta - musin\theta = 0](https://tex.z-dn.net/?f=musin%5Ctheta%20-%20musin%5Ctheta%20%3D%200)
Now, angle between the initial velocities of the objects:
![2\theta = 2\times 70.528^{\circ} = 141.058^{\circ}](https://tex.z-dn.net/?f=2%5Ctheta%20%3D%202%5Ctimes%2070.528%5E%7B%5Ccirc%7D%20%3D%20141.058%5E%7B%5Ccirc%7D)