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3 years ago

How does an object look when it is viewed through a transparent object?

Physics

1 answer:

Vitek1552 [10]3 years ago

5 0

Exactly the same

Explanation:

Objects looks exactly the same when viewed through a transparent object. A transparent body allows light to pass through completely. Therefore, the image formed will completely resemble that of the object to be viewed.

Examples are air, glass, pure water.

A translucent body partially allows light to pass through it. Most times, the image is not clear and it is distorted.

An opaque object will not allow light pass through it. It completely obstructs the ray of path of light. An opaque body will produce an image on the screen.

Learn more:

Refraction brainly.com/question/12370040

#learnwithBrainly

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Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the o

blondinia [14]

The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

$Q=ne$

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

$Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C$

The potential difference V between the disks separated by a distance d is given by,

$V=Ed$

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

$V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V$

The capacitance C of the capacitor is given by,

$C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F$

The capacitance of a parallel plate capacitor is given by,

$C=\frac{\epsilon_0A}{d}$

Here, ε₀ is the permittivity of free space and A is the area of the disks.

Rewrite the expression for A.

$C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3$

the area A of the disks is given by,

$A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }$

Here, D is the diameter of the disk.

$D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm$

The diameter of each disc is found to be 1.32 cm.

7 0

3 years ago

A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w

Sati [7]

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

R² = 2.4

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

3 0

3 years ago

A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s

vovikov84 [41]

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

8 0

3 years ago

1. Biodiversity refers to the variety of life in an ecosystem.<br><br>False<br>True

GalinKa [24]

True
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5 0

3 years ago

A relaxed spring of length 0. 13 m stands vertically on the floor; its stiffness is 1180 N/m. You release a block of mass 0. 5 k

Daniel [21]

The length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m.The length of the spring by the letter x.

<h3>What is the potential energy of the spring?</h3>

The energy is stored in the spring when it is stretched or compressed by some length. It is the product of mass, gravity and distance compressed or stretched. Mathrmatically it is given by;

PE=mgh

The given data in the problem is ;

m is the mass of the block is 0.5 kg height,

h is the height is released is 0.7 m

x initial length of the spring = 0.13 m.

K is the force constant of the spring = 1180 N/m.

By the law of conservation of energy,

The potential energy of the spring gets converted

$\rm mgh=\frac{1}{2}KL^2 \\\\ \rm L= \sqrt{\frac{2mgh}{K} } \\\\ \rm L= \sqrt{\frac{2\times 0.5\times9.81 \times 0.7}{1180} \\\\ \\\\$

$\rm L= 0.054m$

Hence the length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m

To learn more about the potential energy of the spring refer to the link;

brainly.com/question/2730954

4 0

2 years ago