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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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A motorboat traveling with a current can go 160 km in 4 hours. against the current it takes 5 hours to go the same distance. Fin
MatroZZZ [7]
<h2>Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>

Explanation:

Let speed of motor boat be m and speed of current be c.

A motorboat traveling with a current can go 160 km in 4 hours.

   Distance = 160 km

   Time = 4 hours

    Speed = m + c

   We have

            Distance = Speed x Time

            160 = (m+c) x 4

            m + c = 40     --------------------- eqn 1

Against the current it takes 5 hours to go the same distance.

   Distance = 160 km

   Time = 5 hours

    Speed = m - c

   We have

            Distance = Speed x Time

            160 = (m-c) x 5

            m - c = 32     --------------------- eqn 2

eqn 1 + eqn 2

           2m = 40 + 32

             m = 36 km/hr

Substituting in eqn 1

               36 + c = 40

                      c = 4 km/hr

Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.

3 0
3 years ago
In the drawing, water flows from a wide section of a pipe to a narrow section. In which part of the pipe is the volume flow rate
guapka [62]

Answer:

Volume flow rate is the same in both sections of the pipe

Explanation:

7 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
NIST 800-14's Principles for Securing Information Technology Systems, can be used to make sure the needed key elements of a succ
mariarad [96]

Answer:

True

Explanation:

-NIST 800-14's are generally accepted principles for securing information technology systems.

- The defined principles, if adhered to and continously improved, will will ensure sytem security over it's lifetime as desired.

5 0
3 years ago
A bar magnet is cut in half, as shown.
Feliz [49]

Answer:

NS/NS

Explanation:

If we cut a bar magnet in half pieces according to the image that is given below the question, then the magnet will show NS/NS that is (North-South/North-South) because the smaller pieces of the magnet will become a small magnet with the same property-carrying both the poles, north pole as well as south pole. The poles will not be separated.

5 0
3 years ago
Read 2 more answers
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