The force of attraction between two objects can be illustrated using Newton's Law of Universal Gravitation.
The relation between the different parameters is shown in the attached image.
Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.
This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.
Answer:
Explanation:
Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.
Answer: 0°
Explanation:
Step 1: Squaring the given equation and simplifying it
Let θ be the angle between a and b.
Given: a+b=c
Squaring on both sides:
... (a+b) . (a+b) = c.c
> |a|² + |b|² + 2(a.b) = |c|²
> |a|² + |b|² + 2|a| |b| cos 0 = |c|²
a.b = |a| |b| cos 0]
We are also given;
|a+|b| = |c|
Squaring above equation
> |a|² + |b|² + 2|a| |b| = |c|²
Step 2: Comparing the equations:
Comparing eq( insert: small n)(1) and (2)
We get, cos 0 = 1
> 0 = 0°
Final answer: 0°
[Reminders: every letters in here has an arrow above on it]
The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft
The addition of vectors and the uniform motion allows to find the answers for the questions about distance and time are:
- The distance to go between airports A and C is 373.6 10³ m
- The time to go from airport A to B is 2117 s
Vectors are quantities that have modulus and direction, so their addition must be done using vector algebra.
In this case the plane flies towards the North a distance of y = 360 10³ m at an average speed of v = 170 m / s, when arriving at airport B it turns towards the East and travels from x = 100 10³ m, until' it the distance reaches the airport C
Let's use the Pythagoras theorem to find the distance traveled
R = Ra x² + y²
R = 10³
R = 373.6 10³ m
They indicate the average speed for which we can use the uniform motion ratio
v = 
t = 
They ask for the time in in from airport A to B, we calculate
t = 360 10 ^ 3/170
t = 2.117 10³ s
In conclusion we use the addition of vectors the uniform motion we can find the answer for the question of distance and time are:
- The distance to go between airports A and C B is 373.6 10³ m
- The time to go from airport A to B is 2117 s
Learn more here: brainly.com/question/15074838
