The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
Answer:
very small solid particles called interstellar dust.
Explanation:
In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.
Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).
Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.
Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.
Answer:
a) 200A
b) 10.2V
c) 2.04kW
d)
I=80A
V=4.08V
P=0.326kW
Explanation:
Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:
Where R is the equivalent resistance of the resistors in series
![R=0.0510+0.0090=0.0600[ohm]](https://tex.z-dn.net/?f=R%3D0.0510%2B0.0090%3D0.0600%5Bohm%5D)
To calculate the voltage dropped by the motor we have to apply the voltage divider rule:
The power dissipated supplied to the motor is given by:
now solving adding a 0.0900 ohm resistor:
Directly proportional to pressure
Answer:
q = 400 nC
the correct answer is b
Explanation:
The expression for the electric potential of a point charge is
V = k q / r
they ask us for the electrical charge
q = V r / k
let's calculate
Q = 600 6.0 / 9 10⁹
Q = 4 10⁻⁷ C
let's reduce to nC
Q = 4 10⁻⁷ C (10⁹ nC / 1C)
q = 4 10² nC = 400 nC
the correct answer is b
Traslate
La expresión para el potencial eléctrico de una carga puntual es
V = k q/r
nos piden la carga eléctrica
q= V r /k
calculemos
Q= 600 6,0 / 9 10⁹
Q= 4 10⁻⁷ C
reduzcamos a nC
Q = 4 10⁻⁷ C(10⁹ nC/1C )
q = 4 10² nC = 400 nC
la respuesta correcta es b
