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Anuta_ua [19.1K]

2 years ago

11

An 8.0-kg history textbook is placed on a 1.25-m high desk. How large is the gravitational potential energy of the textbook-Eart

h system relative to the desktop?

2 answers:

denpristay [2]2 years ago

5 0

<h2>Answer</h2>

Gravitational potential energy of the textbook-Earth system relative to the desk = 98 N/m

<h2>Explanation</h2>

Given that,

Mass of history book = 8kg

height of book from ground = 1.25m

acceleration due to gravity = 9.8m/s²

<h3>gravitational potential energy formula</h3><h2>GPE = Fg⋅h</h2><h2>GPE = mgh</h2>

where ,

m = mass of object(book)

g = acceleration due to gravity

h = the altitude of the object.

GPE = 8(9.8)(1.25)

= 98 N/m

So, gravitational potential energy of the textbook-Earth system relative to the desk = 98 N/m

Nesterboy [21]2 years ago

4 0

Answer: 98 J

Explanation: The energy possessed by a body due to its placement in gravitational field. It is given by:

G.P.E. = m g h

m is the mass of the object, g is the acceleration due to gravity and h is the height.

mass of the book, m = 8.0 kg

height of the desk on which book is kept, h = 1.25 m

acceleration due to gravity, g = 9.8 m/s²

G.P.E = 8.0 kg × 9.8 m/s² × 1.25 m = 98 J.

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one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?

Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

(150 horsepower) x (746 watt/HP) x (1 joule/sec / watt) x (10 sec)

= (150 x 746 x 1 x 10) joule = 1,119,000 joules .
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8 0

1 year ago

A student lifts a set of books off a table and places them in the upper shelf of a book case which is 2 meters above the table.

AfilCa [17]

The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle $\theta$ between the direction of the force and the direction of the displacement:
$W=Fd \cos \theta$
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so $\cos \theta = \cos 0=1$
Therefore, the work done is
$W=Fd=(5 N)(2 m)=10 J$

6 0

2 years ago

Read 2 more answers

Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper

vladimir1956 [14]

Answer:

Total distance does the grasshopper travel before the cars hit is 150 m

Explanation:

Each car moves x=100 m before they collide. Both the cars moving in constant velocity. time taken t by each car is

$t=\frac{x}{v}$

where x is the distance traveled with velocity v

$t=\frac{100}{10}\\t=10 sec$

The insect is moving through this time period with a constant velocity of 15 m/s

The distance traveled by grasshopper is

$distance=V_{gh} \times t\\distance=15 \times 10\\distance=150 m$

7 0

2 years ago

he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th

AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x $10^{-6}$ T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ $_{0}$ I/2πr

where

B = magnetic field (T)

μ $_{0}$ = 1.26 x $10^{-6}$ (N/ $A^{2}$ )

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.19)

= 4.96 x $10^{-6}$ T

BV = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.41)

= 2.3 x $10^{-6}$ T

hence,

the net magnetic field = BH - BV

= 4.96 x $10^{-6}$ - 2.3 x $10^{-6}$

= 2.66 x $10^{-6}$ T

4 0

1 year ago

It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging riv

zzz [600]

Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

$V_2^2=V_o^2-2*g*\Delta h$

$V_2^2=-2*10*(-31)$

$V_2=24.9m/s$ This will be our initial speed for a balance of energy.

By conservation of energy:

$m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0$

Where

$h$ is your height at its maximum elongation

$h_o$ is the height of the bridge

$l_o$ is the length of the unstretched bungee cord

$800h+21*(79-h)^2-63200-24800.4=0$

$21h^2-2518h+43060.6=0$ Solving for h:

$h_1=20.66m$ and $h_2=99.24m$ Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

6 0

2 years ago

Read 2 more answers