Yes,it's true ok? So how have you been doing
Answer:
the answer is 5 electrons
Explanation:
because its the same name as the amount of protons
Answer:
d. We can calculate it by applying Newton's version of Kepler's third law
Explanation:
The measurements of a Star like the Sun have several problems, the first one is distance, but the most important is the temperature since as we get closer all the instruments will melt. This is why all measurements must be indirect because of the effects that these variables create on nearby bodies.
Kepler's laws are deduced from Newton's law of universal gravitation, in these laws the mass of the Sun affects the orbit of the planets since it creates a force of attraction, if measured the orbit and the time it takes to travel it we can know the centripetal acceleration and with it knows the force, from where we clear the mass of the son.
Let's review the statements of the exercise
.a) False. We don't have good enough models for this calculation
.b) False. The size of the sun is very difficult to measure because it is a mass of gas, in addition the density changes strongly with depth
.c) False. The amount of light that comes out of the sun is not all the light produced and is due to quantum effects where the mass of the sun is not taken into account
.d) True. This method has been used to calculate the mass of the sun and the other planets since the variable distance and time are easily measured from Earth
Correct answer is D
The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW
Answer:
(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East
(b) When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West
Explanation:
Given;
a force vector points due east,
= 140 N
let the second force = 
let the resultant of the two vectors = F
(a) When the resultant force is pointing along east line
the second force must be pointing due east


(b) When the resultant force is pointing along west line
the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

