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matrenka [14]
3 years ago
5

How much force do you push down on the Earth with? (Hint: 1kg is about 2 pounds).​

Physics
2 answers:
e-lub [12.9K]3 years ago
6 0

The force with which I push down on the Earth is PRECISELY equal to the force with which the Earth pushes UP on me.  

I weigh EXACTLY the same amount on the Earth as the Earth weighs on me.  Find THAT amount, and you'll have the answer to the question.

(Hint:  1 kg is a mass, and 2 pounds is a force, so they can never be equal.  1kg weighs about 2.2 pounds IF it happens to be on Earth.  But if the same 1kg is on the Moon, then it weighs about 5.8 ounces.  And if it happens to be on Mars, then it weighs about 13.3 ounces.  The same kilogram weighs different weights in different places.)

On a good day, I push down on the Earth with about 890 Newtons of force.  No matter how hard I try, I can never push more or less than that.

Gennadij [26K]3 years ago
3 0

Answer: 100lbs

Explanation: The Earth pushes you down at 100lbs, so you push down the earth by 100lbs which is enough to keep you firmly attached to the ground and not allow you to jump more than a couple of feet.

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Because Daltons theory was more in-depth and persist then Democritus.
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1. Have you ever tried to undergo an X-ray to test on your body? What can you see in the fils aftee the examination? why is the
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when an object is immersed in liquid at rest,why is the net force on the object is the horizontal direction equal to zero?​
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6 0
3 years ago
A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?
ra1l [238]

Answer:

a=-.78m/s^2

Explanation:

Δv=at

  • Δv is the difference in velocity before and after a given time.
  • a is the acceleration of the object during this time.
  • t is time

(v_f-v_i)=at is another way to write this equation.

  • The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δv=(v_f-v_i)

v_f-v_i=at\\\frac{at}{t}=\frac{v_f-v_i}{t}\\a=\frac{v_f-v_i}{t}

  • I rearranged this equation to solve for a, but this is a step that you don't need to take, it's just good to get in the habit of doing this.
  • Plug in the given values. Note that our final velocity is 0, because the car travels until at <em>rest</em>.

a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}

  • Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}

  • if you converted correctly, your answer for v_f will be ≅ 7.6m/s.
  • Now divide. Notice that the units for acceleration are m/s^2 or <em>meters per second, per second</em>.

a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2

  • Our final answer is <em>negative </em>because the car is <em>slowing down</em>. Do not square this answer as the square symbol only applies to the units, not the magnitude.
4 0
3 years ago
Read 2 more answers
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Ludmilka [50]

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We have density of gasoline = 0.77 kg/L = 6.073 lb/US gal

Mass rate = Density * Volume rate

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So mass flow rate delivered by the gasoline pump in lbm/min = 56.68

6 0
4 years ago
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