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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?
Physics
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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A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the
leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B )  , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

=  - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

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So magnetic field B = .28 i + .115 B₃

Part A

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Part B

y component of B = 0

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8 0
2 years ago
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

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(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

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5 0
2 years ago
A rope breaks when the tension reaches 205 N. What is the maximum speed at which it can swing a 0.477 kg mass in a circle of rad
jekas [21]

Answer:

Maximum velocity will be 17.651 m /sec

Explanation:

We have given a rope breaks when tension reaches 205 N

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8 0
2 years ago
Which of the following best describes a hypothesis?
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I am 99% sure it is B :)
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