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2 years ago

The earths gravity is pulling on you, Are you pulling on the earth Explain your reasoning with evidence

1 answer:

5 0

Yes, you are pulling on the Earth, with equal force.
Your weight on Earth is equal to the Earth's weight on you.

I can't think of a way for you to go out and see it or measure it,
but there are two things we learn in Physics that tell us that it
must be true.

#1). Newton's 3rd Law of motion:
For every action, there is an equal, opposite reaction.
If the Earth pulls on me, then I must pull on the Earth with an
equal, opposite force.

#2). Newton's law of universal gravitation:

Gravitational force = (a Constant) · (mass₁) · (mass₂) / (distance)² .

Take a good, long, hard look at this formula for the gravitational force.

It tells the strength of the gravitational force between the Earth and you.
But it doesn't say WHICH object is (mass₁) and which object is (mass₂) !
It doesn't matter ! BOTH objects feel the same gravitational force !

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nsider a common mirage formed by superheated air immediately above a roadway. A truck driver whose eyes are 2.00 m above the roa

Anna11 [10]

Answer:

$n_{air} =1.0000355$

Explanation:

from snells law we have following relation

$n' Sin i = n_{air} Sin r$

$
n_{air}$ is refractive index of air

n' - refractive index of air below

i is angle of incidence = 90 - 1.30 = 88.70^o

r is refraction angle

plugging all value in above formula

$1.000293 \times sin 88.7 = n_{air} \times sin 90$

$n_{air} =1.0000355$

6 0

3 years ago

Plz HELP! All of the following are possible reasons to complete strength training (anaerobic) exercises, except:

Illusion [34]

Answer:i would say "to treat or prevent injuries"

Explanation: anaerobic exercise is: "high intensity movements preformed in a short period of time."

to treat or prevent injuries you would want to do Aerobic exercise, which is an easy exercise like walking.

4 0

3 years ago

Read 2 more answers

A car starts from rest at a stop sign. It accelerates at 4.0 m/s2 for 3 seconds, coasts for 4 s, and then slows down at a rate o

solmaris [256]

Well, Break the problem up into parts. For speeding up: The car accelerates at 4 m/s^2 for 3 seconds. Multiplying these values tells you the car reaches a speed of 12 m/s. Vf^2 = Vi^2 + 2a(Xf - Xi) 12^2 = 0 + 2(4)(Xf - 0) 144 = 8 Xf Xf = 18 m For coasting: The car is at the 12 m/s and does this for 2 seconds. x = vt = (12)(2) = 24 m For slowing down: The car decelerates at 3 m/s^2 to come to a stop at the next sign. From 12 m/s, this would take 12/3 seconds or 4 s. Vf^2 = Vi^2 + 2a(Xf - Xi) 0 = 12^2 + (2)(-3)(Xf - 0) -144 = -6 Xf Xf = 24 m Summing the distances: Xtotal = 18 + 24 + 24 = 66 m

3 0

2 years ago

Read 2 more answers

Which of the following pressures rises and falls with the phases of breathing but eventually equalizes with the pressure of the

VikaD [51]

Answer:

(c) Intrapulmonary pressure

Explanation:

Intrapulmonary pressure is the pressure of air within the alveoli, it changes with the different phases of breathing, and because it is connected to the atmosphere through the throat if eventually equalizes with the atmospheric air pressure in the environment.

7 0

2 years ago

A car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another

Alex_Xolod [135]

Answer:

15.8 ft/s

Explanation:

$\frac{da}{dt}$ = Velocity of car A = 9 ft/s

a = Distance car A travels = 21 ft

$\frac{db}{dt}$ = Velocity of car B = 13 ft/s

b = Distance car B travels = ft

c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s$

∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s

6 0

3 years ago