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avanturin [10]
2 years ago
13

Calculate the work done to push a 200-N object 5-meters

Physics
1 answer:
AVprozaik [17]2 years ago
6 0
1000 joules
jskskmxk dnd
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A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The boo
In-s [12.5K]

Answer:

Explanation:

In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.

The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.

Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.

To find the weight of the book we simply multiply the mass of the book by gravity.

W = m*g

W = 1.3[kg] * 9.81[m/s^2]

W = 12.75 [N]

7 0
3 years ago
A badger is running at a speed of 1 m/s. If the badger moves that was for 2600 seconds, how far will the badger travel?
katen-ka-za [31]

Answer:

It would be 2600

Explanation:

M/S stands for meters per second. If it moved 1 meter for 2600 seconds, than it would be 2600. You just multiply 2600 by 1! I hope this helps :D

8 0
3 years ago
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s
IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

r = \frac{a(1-E^2)}{1+Ecos\beta }   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

\beta = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

v^2 = \frac{4\pi^2 }{r_{c} }   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

(V^2 = (\frac{4\pi^{2}  }{149.626*10^9})

therefore : V = 1.624* 10^-5 m/s

4 0
4 years ago
of 72.0 units. If the mass of Object 1 is halved AND the mass of object 2 is tripled, then the new gravitational force will be u
katen-ka-za [31]

Answer:

youtg jugre contact coming call details

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2 years ago
A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of
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Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

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