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Rus_ich [418]
3 years ago
6

A 1.60 m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the

book relative to (a) the ground and (b) the top of the person’s head? How is the work done by the person related to the answers in parts (a) and (b)?
Physics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

Explanation:

Given

mass of book(m)=2.1 kg

height up to which book is lifted is (h)2.2 m

height of person (h_0)1.6 m

Potential energy of book relative to ground=mgh

PE=2.1\times 9.8\times 2.2=45.276 J

(b)PE w.r.t to person head =mg(h-h0)

=2.1\times 9.8\times (2.2-1.6)=12.348 J

work done by person in lifting box 2.2 m w.r.t floor

Word done =Potential Energy of box relative to floor=45.2 J

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Gravity anything with mass as gravity we know the earth has gravity because you and I are standing on earth and not floating off
Paraphin [41]

Answer:

F=mg

Explanation:

Close to Earth's surface, the force of gravity that pulls an object towards the ground is

F=mg (2)

where

m is the mass of the object

g is the acceleration due to gravity, which is 9.81 m/s^2 close to Earth's surface

This is an approximation of the general formula of gravity valid only close to Earth's surface. The more general formula is

F=G\frac{Mm}{r^2} (1)

where

G is the gravitational constant

M is the Earth's mass

m is the object's mass

r is the distance of the object from Earth's center

At the Earth's surface,

r = R (Earth's radius), and by calling the following factor

g=\frac{GM}{R^2}

we see that eq.(1) becomes eq.(2).

8 0
2 years ago
A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally
leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have

<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>

where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

3 0
3 years ago
What is the mass of an object travelling at 25 m/s with a kinetic energy of 3775 J?
Arte-miy333 [17]

Answer:

<h2>You can do 6516-3775=2741</h2><h2 /><h2>The difference is 27141 kilometers</h2><h2 /><h2>Did this help?</h2><h2>⇒  Yes or No?</h2>
4 0
2 years ago
A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri
s344n2d4d5 [400]

Answer:

The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad

Explanation:

8 0
2 years ago
PLSSS HELP MEEEE
vekshin1

Answer:

Cu2+

Explanation:

7 0
3 years ago
Read 2 more answers
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