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3 years ago

14

A coin is placed at a depth of 15 cm in a beaker from the surface of water. Therefractive index of water is 4/3.Calculate height

through which the image of the coin is raised.

1 answer:

Whitepunk [10]3 years ago

4 0

I found this answer on another website.

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Two wires of the same material and having the same volume, are fixed

Setler79 [48]

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ( $r_{1}$ )² × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (( $r_{2}$ )² ÷ 4) × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$ )

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ( $m_{1}$ × g) ÷ ( $A_{1}$ )

Stress in second wire = ( $m_{2}$ × g) ÷ ( $A_{2}$ ) = ( $m_{2}$ × g) ÷ (4 × $A_{1}$ )

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (( $m_{1}$ × g) ÷ ( $A_{1}$ )) ÷ (ΔL ÷ (4 × $l_{2}$ ))

Stress per unit strain of second wire = (( $m_{2}$ × g) ÷ (4 × $A_{1}$ )) ÷ (ΔL ÷ $l_{2}$ )

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg

5 0

3 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

4 years ago

Which body part is included in circulatory system

velikii [3]

The heart and the blood vessels.

6 0

4 years ago

Find the work done when a constant force f = 13 lbs moves a chair from x = 2.4 to x = 4.1 ft. along the x -axis.

Blizzard [7]

The work done in moving the chair from x = 2.4 to x = 4.1 ft using a force of F = 13 lbs is 29.98 J

<h3>What is work done? </h3>

This is simply defined as the product of force and distance moved in the direction of the force. Mathematically, it can be expressed as

Workdone (Wd) = force (F) × distance (d)

Wd = Fd

<h3>How to convert lb to N</h3>

1 lb = 4.45 N

Therefore,

13 lbs = 13 × 4.45

13 lbs = 57.85 N

<h3>How to determine the distance in m</h3>

Initial distance = 2.4 ft
Final distance = 4.1 ft
Change in distance = 4.1 -2.4 = 1.7 ft

1 ft = 0.3048 m

Therefore,

1.7 ft = 1.7 × 0.3048

1.7 ft = 0.51816 m

<h3>How to determine the workdone</h3>

Force (F) = 57.85 N
Distance (d) = 0.51816 m
Workdone (Wd) =?

Wd = Fd

Wd = 57.85 × 0.51816

Workdone = 29.98 J

Thus, the workdone in moving the chair is 29.98 J

Learn more about workdone:

brainly.com/question/17358222

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3 0

2 years ago

Which is a way that biotechnology has Not helped society?

QveST [7]

Bacteria can be used to grow a lot of vaccines. new, stronger bacteria has evolved that are resistant to antibiotics. Diabetics have a steady supply of inexpensive insulin

4 0

3 years ago