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Delvig [45]
3 years ago
5

How does universal gravitation affect you and objects around you?

Physics
1 answer:
Olegator [25]3 years ago
8 0

<u>Answer</u>:

Newton proposed the "law of Universal Gravitation". He said that gravity is universal in nature and it affects all the objects. Gravity is present between you and every object having mass around you. The "Strength of the gravity" between two objects with mass is depends on the product of masses and distance between the objects.

The "Gravitational force" increases with the increase in mass of the objects and decreases with the increase in distance between the objects. The gravity exerts a force on every object. The direction of force of gravity is towards the centre of the Earth.  

 

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Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 20 mph and half the d
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Answer:

On the way to grandmother´s, the average speed was 30 mph. On the way back, the average speed was 40 mph.

Explanation:

The average speed is given by the variation of the position over time.

Mathematically:

ΔX / Δt = v

where:

ΔX = distance (final position - initial position)

Δt = time (final time - initial time)

v = speed

On the way to Grandmother´s, we can calculate how much time Julie drove at each speed:

ΔX / Δt = v

ΔX / v = Δt

50 mi / 20 mph = 2.5 h

In the same way, we can calculate how much time she drove at 60 mph:

50 mi / 60 mph = 0.83 h

In total, she drove a distance of 100 mi in (2.5 h + 0.83 h) 3.33 h. Then, the average speed on the way to Grandmother´s was:

<u>ΔX / Δt = v = 100 mi / 3.33 h = 30 mph</u>

In the return trip, we do not know the distance nor the time that she drove at each speed, but we know that for each part of the trip, the time is the same (Δt)  and we also know that the total distance is 100 mi and the total time is 2Δt:

v1 = ΔX1 / Δt

v2 = ΔX2 / Δt

ΔX2 + ΔX1  = 100

where

v1 = speed during the first part of the trip (20 mph)

v2 = speed during the second part of the trip (60 mph)

ΔX1 = distance driven at 20 mph

ΔX2 = distance driven at 60 mph

Δt = time

If we divide v2/v1, we will get:

v2/v1 = (ΔX2 / Δt) / (ΔX1 / Δt)

60 mph / 20 mph = ΔX2 / ΔX1

3 = ΔX2 / ΔX1

3ΔX1 = ΔX2

Then we can replace ΔX2 for 3ΔX1 in this equation:

ΔX2 + ΔX1  = 100 mi

3ΔX1 + ΔX1 = 100 mi

4ΔX1 = 100 mi

ΔX1 = 25 mi

And now, we can solve Δt from the equation of v1:

v1 = ΔX1 / Δt

Δt = ΔX1 / v1 = 25 mi / 20 mph = 1.25 h

The average speed on the return trip is then:

<u>v = 100 mi / 2Δt = 100 mi / 2.5 h = 40mph</u>

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