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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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Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions about a fixed axis. Compare the linear
dezoksy [38]

Answer:

Explanation:

Wheel completes four revolution.

The linear displacement is zero.

The angular displacement is 4 x 2π = 8π radian.

So, option (c) is correct.

4 0
2 years ago
a film of transparent material 120 nm thick and having refractive index 1.25 is placed on a glass sheet having refractive index
Eva8 [605]

Answer:

a) 600nm

b) 300nm

Explanation:

the path difference = 2t  

t = thickness of the film

L' = wavelength of light in film = L/n

L = wavength of light in air

n = refractive index of glass

(a)

for destructive interference 2t = L'/2 = L/2n

L = 4*t*n

= 4*120*10^-9*1.25  

L = 600 nm

(b)

for constructive interference 2t = L' = L/1.25

L = 2tn

= 2 × 1.25 ×  120nm

= 300 nm

4 0
2 years ago
Sunlight strikes a piece of crown glass at an angle of incidence of 38.0°. Calculate the difference in the angle of refraction b
zhuklara [117]

Answer:

Difference in the angle of refraction = 0.3°

41.04° is the minimum angle of incidence.

Explanation:

Angle of incidence  = 38.0°

For yellow light :

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for yellow light which is 1.523

n₂ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.523}{1}

{sin\theta_2}=0.9377

Angle of refraction for yellow light = sin⁻¹ 0.9377 = 69.67°.

For green light :

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for green light which is 1.526

n₂ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.526}{1}

{sin\theta_2}=0.9395

Angle of refraction for green light = sin⁻¹ 0.9395 = 69.97°.

The difference in the angle of refraction = 69.97° - 69.67° = 0.3°

Calculation of the critical angle for the yellow light for the total internal reflection to occur :

The formula for the critical angle is:

{sin\theta_{critical}}=\frac {n_r}{n_i}

Where,  

{\theta_{critical}} is the critical angle

n_r is the refractive index of the refractive medium.

n_i is the refractive index of the incident medium.

n₁ is the refractive index for yellow light which is 1.523 (incident medium)  

n₂ is the refractive index of air which is 1 (refractive medium)

Applying in the formula as:

{sin\theta_{critical}}=\frac {1}{1.523}

The critical angle is = sin⁻¹ 0.6566 = 41.04°

5 0
2 years ago
Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w
Komok [63]
There is a ner force of 15 N allowing Levi and Clara to mobe the box.
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Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x
never [62]

Answer:

210.3 degrees

Explanation:

The net force exerted on charge A = 59.5 N

Use the x and y coordinates of net force to get the direction

arctan (y/x)

8 0
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