Determine the final angular velocity of a particle that rotates 4500 ° in 3 seconds and an angular acceleration of 8 Rad / s ^ 2

Answer: her rotational kinetic energy increases

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.

For this "type" of motion, displacement (Δx) can be determined by:

[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]

[tex]v_{i}[/tex] is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]

[tex]\Delta x = 30 - 2.36[/tex]

[tex]\Delta x[/tex] = - 42

Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.

Answer:

a)  λ = 555.5 m,  λ = 187.5 nm

Explanation:

The velocity of a wave is given by the relation

           c = λ f

           λ = c /f

a) length of the radii AM

            λ = 3 10⁸/540 10⁻⁷

            λ= 5.555 102 m

            λ = 555.5 m

             f = 1600 kHz

            λ = 3 108/1600 103

             λ = 1.875 102 m

            lam  = 187.5 nm

b) light = 760 Thz = 760 10-12 Hz

         λ=  c/f

            λ = 3 108/760 10-12

            λ = 3.947 10-9

           λ= 30000 Thz

           λ= c/f

      λ = 3 10⁸/ 30000 10-12

       λ = 1  m

Answer:

The mass of the object is 3.08 kg.

Explanation:

The horizontal force is12.7 N and the coefficient of the kinetic fraction are 0.42. Now we have to compute the mass of the object. Thus, use the below formula to find the mass of the object.

Let the mass of the object = m.

The coefficient of kinetic friction, n = 0.42

Therefore,  

Force, F = n × mg

12.7 = 0.42 × 9.8 × m

m = 3.08 kg

The mass of the object is 3.08 kg.

Answer:

When the entire liquid is a single temperature

Explanation:

When a liquid is heated, a convection current is set up. Convection is the movement of

fluid particles in response to a temperature gradient.

When you start heating a liquid, the particles near the base of the heating vessel increase in temperature, become less dense and rise upwards while the denser particles move downwards. This convection current will continue until an equilibrium temperature is obtained throughout the liquid.

Answer:

The wave is travelling in the ±z-axis direction.

Explanation:

An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the  electric and magnetic field.

In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.

Answer:

Explanation:

We shall apply Brewster's law to solve the problem . Let i be the angle of reflection required. According to this law ,

tan i = μ where i is angle of incidence , μ is refractive index of the medium .

Here i = ? , μ = 1.31

tani = 1.31

i = 53° approx .

At the angle of 53° incidence , the reflected ray will be completely polarised .

Answer:

2.29e-9C/m²

Explanation:

Using E = σ/ε₀ means the force on the electron is F = eE = eσ/ε₀.

The work done on the electron is W = Fd = deσ/ε₀. This equals the kinetic energy lost, ½mv².

½mv² = deσ/ε₀

d = 75cm – 15cm = 60cm = 0.6m

σ = mv²ε₀/(2de)

. .= 9.11e-31 * (7.4e6)² * 8.85e-12 / (2 * 0.6 * 1.6e-19)

. .= 2.29e-9 C/m² (i.e. 2.29x10^-9 C/m²)

Answer: f ≈ 8.5Hz

Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.

For a source moving and a stationary observer, to determine the frequency:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

where:

[tex]f_{0}[/tex] is frequency of observer;

[tex]f_{s}[/tex] is frequency of source;

c is the constant speed of sound c = 340m/s;

[tex]v_{s}[/tex] is velocity of source;

Rearraging for frequency of source:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]

Replacing and calculating:

[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]

[tex]f_{s} = 9.(0.9412)[/tex]

[tex]f_{s} =[/tex] 8.5

Frequency the horns emit is 8.5Hz.

Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

Q= required charge

Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

diameter of the conductor is 4.20 mm

But Radius= diameter/2= 4.20/2=

The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then

radius of the conductor is 0.0021m.

We can now calculate the area of the conductor which is

A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

We can proceed to calculate the current density below

J = 140/13.85*10^-6 = 10108303A/m

According to the listed reference:

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s

Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

(a) The charge passes through the starter motor is 106.4C.

(b) An electron travel along the wire while the starter motor is on 0.5676mm.

Answer (a)

I= current

t= time taken

Q= required charge

Q= I*t

Q= 140*0.760

Q= 106.C

Answer (b)

The n electron travel along the wire while the starter motor is on:

Diameter of the conductor is 4.20 mm

Radius= diameter/2= 4.20/2

Radius =2.1mm

Radius of the conductor is 0.0021m.

A = π*r^2

A= π*(0.0021)^2

A= 13.85*10^-6 m^2

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )

Vd  =0.0007468m/s

Vd =0 .7468 mm/s

The distance traveled is:

x = v*t

x= 0.7468 * 0.760

x = 0.5676mm

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Answer:

The radius of the black hole will be 2949.6 m.

Explanation:

The radius of this black hole will be the Schwarzschild radius of the mass of the sun

[tex]r_{s}[/tex] = [tex]\frac{2GM}{c^{2} }[/tex]

where

G is the gravitational constant = 6.67 x 10^-11 m^3⋅kg^-1⋅s^-2

M is the mass of the sun = 1.99×10^30 kg

c is the speed of light = 3 x 10^8 m/s

substituting values into the equation, we have

[tex]r_{s}[/tex] = [tex]\frac{2*6.67*10^{-11}*1.99*10^{30} }{(3*10^{8} )^{2} }[/tex] = 2949.6 m

Complete question:

A solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of 6.75 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

Answer:

(a) the energy density of the magnetic field inside the solenoid is 50.53 J/m³

(b) the total energy stored in the magnetic field is 0.121 J

Explanation:

Given;

length of the solenoid, L = 98.6 cm = 0.986 m

cross-sectional area of the solenoid, A = 24.3 cm² = 24.3 x 10⁻⁴ m²

number of turns of the solenoid, N = 1310 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μ₀nI

B = μ₀(N/L)I

Where;

μ₀ is permeability of free space, = 4π x 10⁻⁷ m/A

[tex]B = \frac{4\pi*10^{-7}*1310*6.75}{0.986} \\\\B = 0.01127 \ T[/tex]

(a) Calculate the energy density of the magnetic field inside the solenoid

[tex]u = \frac{B^2}{2 \mu_o}\\\\u = \frac{(0.01127)^2}{2*4\pi *10^{-7}} \\\\u = 50.53 \ J/m^3[/tex]

(b) Find the total energy stored in the magnetic field

U = uV

U = u (AL)

U = 50.53 (24.3 x 10⁻⁴  x 0.986)

U = 0.121 J

Answer:

A) 0.1477

B) 0.65388 mm

C) object is inverted

Explanation:

The formula for object - image relationships for spherical reflecting surface is given as;

n1/s + n2/s' = = (n2 - n1)/R

Where;

n1 & n2 are the Refractive index of both surfaces

s is the object distance from the vertex of the spherical surface

s' is the image distance from the vertex of the spherical surface

R is the radius of the spherical surface

We are given;

index of refraction of glass; n2 = 1.60

s = 24 cm = 0.24 m

R = 4 cm = 0.04 m

index of refraction of air has a standard value of 1. Thus; n1 = 1

a) So, making s' the subject from the initial equation, we have;

s' = n2/[((n2 - n1)/R) - n1/s]

Plugging in the relevant values, we have;

s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]

s' = 0.1477

b) The formula for lateral magnification of spherical reflecting surfaces is;

m = -(n1 × s')/(n2 × s) = y'/y

Where;

m is the magnification

n1, n2, s & s' remain as earlier explained

y is the height of the object

y' is the height of the image

Making y' the subject, we have;

y' = -(n1 × s' × y)/(n2 × s)

We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.

Thus;

y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)

y' = - 0.00065388021 m = -0.65388 mm

C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.

Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.

Thus, the object is inverted since m is negative.

Answer:

se the principle of induction.

place the two metallic spheres together,  now we bring the positively charged bar closer to the first sphere.

The charge that was induced in the sphere is distributed as infirm as possible,

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive

Explanation:

For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.

Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.

Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.

The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.

Answer:

Explanation:

As temperature is constant , we shall apply Boyle's law

P₁V₁ = P₂V₂

P₁ = pressure at depth of 10 m

= P + hdg , h = 10 , d = 10³ , g = 10

P is atmospheric pressure which is 10⁵ Pa

P₁ = 10⁵ + 10 x 10³ x 10

= 2 x 10⁵

applying the formula

2 x 10⁵ x 6 = 10⁵ x v

v = 2 x 6 = 12 L

volume will be doubled at the surface .

B )

warming of air at the surface will increase the volume of air in her lungs so so she will need more lung capacity .

C )

The rms value of a gas depends upon the temperature of the gas . As temperature of the gas is constant , the rms value of the gas particles will remain constant when she goes to the surface .

The lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.

According to the law, the pressure of the gas is inversely proportional to the volume of the gas. In other words when the pressure of the gas increases the volume of the gas decreases.

We know the pressure at the 10 meters depth:

[tex]\rm P_1 = P+h\times \rho\times g[/tex]

Where P = Atmospheric pressure

            h = Depth

            ρ =Density of the water

We have: [tex]\rm P = 10^5 \ Pa[/tex], h = 10 meters, and [tex]\rm \rho = 1000 \ kg/m^3[/tex], and [tex]\rm g = 10 \ m/s^2[/tex]

Putting the values in the above equation, we get:

[tex]\rm P_1 = 10^5+ 10\times 1000\times 10[/tex]

[tex]\rm P_1 = 2\times 10^5[/tex]

From the Boyle's law:

[tex]\rm P_1\times V_1 = P_2\times V_2[/tex]

[tex]\rm 2\times10^5\times 6 = 10^5\times V_2[/tex]

[tex]\rm V_2 = 12 \ L[/tex]

We know that as the air at the surface warms, the volume of air in her lungs expands, requiring more lung capacity.

The temperature of the gas is constant and [tex]\rm V_{rms}[/tex] values for gas depend on the temperature of the gas, but here the temperature of the gas is constant thus, the  [tex]\rm V_{rms}[/tex] will remains constant.

Thus, the lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.

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Answer:

s = 2.16 x 10¹¹ m

Explanation:

Since, the waves travelling from Earth to the Mars rover are electromagnetic. Therefore, there speed must be equal to the speed of light. So, from the equation given below:

s = vt

where,

s = the distance between Earth and Mars = ?

v = speed of the wave = speed of light = 3 x 10⁸ m/s

t = time taken by the radio signals to reach the rover from Earth

t = (12 min)(60 s/1 min) = 720 s

Therefore,

s = (3 x 10⁸ m/s)(720 s)

s = 2.16 x 10¹¹ m

Answer:

a) 0.67 rad/sec in the clockwise direction.

b) 98.8% of the kinetic energy is lost.

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia [tex]I[/tex] = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia [tex]I[/tex] = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = [tex]I[/tex]ω

where [tex]I[/tex] is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = [tex]I[/tex]ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = [tex]I[/tex]ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s

When they are coupled coupled together, their total rotational inertia [tex]I_{t}[/tex] = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = [tex]I_{t}[/tex][tex]w_{f}[/tex]

where [tex]I_{t}[/tex] is their total rotational inertia

[tex]w_{f}[/tex] = their final angular speed together

Final angular momentum = 3 x [tex]w_{f}[/tex] = 3[tex]w_{f}[/tex]

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 =  3[tex]w_{f}[/tex]

[tex]w_{f}[/tex] = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s

From the first statement, the direction is clockwise

b) Rotational kinetic energy = [tex]\frac{1}{2} Iw^{2}[/tex]

where [tex]I[/tex] is the rotational inertia

[tex]w[/tex] is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = [tex]\frac{1}{2} Iw^{2}[/tex] = [tex]\frac{1}{2}*2*5^{2}[/tex] = 25 J

for second cylinder = [tex]\frac{1}{2}*1*8^{2}[/tex] = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = [tex]\frac{1}{2}I_{t}w^{2} _{f}[/tex]

where

where [tex]I_{t}[/tex] is the total rotational inertia of the cylinders

[tex]w_{f}[/tex] is final total angular speed of the coupled cylinders

Final kinetic energy =  [tex]\frac{1}{2}*3*0.67^{2}[/tex] = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%

A) The angular speed of the combination of the two cylinders is; ω₃ = 0.67 rad/s

B) The percentage of the original kinetic energy lost to friction is;

percentage energy lost = 98.82%

We are given;

Rotational Inertia for first cylinder; I₁ = 2 kg.m²

Angular speed of first cylinder; ω₁ = 5 rad/s

Angular speed of second cylinder; ω₂ = 8 rad/s

Rotational Inertia for second cylinder; I₂ = 1 kg.m²

From conservation of angular momentum, we know that;

Initial angular Momentum([tex]L_{i}[/tex]) = Final angular Momentum([tex]L_{f}[/tex])

Thus;

I₁ω₁ + I₂ω₂ = I₃ω₃

Where;

ω₃ is the angular speed when the two cylinders are combined

I₃ = I₁ + I₂

I₃ = 2 + 1

I₃ = 3 kg.m²

Since the second cylinder rotates in an anticlockwise direction, then its' angular speed will be negative. Thus;

(2 * 5) + (1 * -8) = 3ω₃

10 - 8 = 3ω₃

3ω₃ = 2

ω₃ = 2/3

ω₃ = 0.67 rad/s

B) Let us find initial kinetic energy;

E_i = ¹/₂I₁ω₁² + ¹/₂I₂ω₂²

E_i = ¹/₂((2 * 5²) + (1 * 8²)

E_i = 57 J

Final kinetic energy is;

E_f = ¹/₂I₃ω₃²

E_f = ¹/₂ * 3 * 0.67²

E_f = 0.67335 J

Energy lost = 57 - 0.67335 = 56.32665 J

percentage energy lost = (56.32665/57) * 100%

percentage energy lost = 98.82%

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Answer:

The pressure is [tex]P = 583333 \ N/m^2[/tex]

Explanation:

From the question we are told that

  The area of the edge is  [tex]A = 0.72 cm^2 = 0.72 *10^{-4}\ m[/tex]

    The  force is [tex]F = 42 \ N[/tex]

The pressure is mathematically represented as

            [tex]P = \frac{F}{A}[/tex]

substituting values

           [tex]P = \frac{42}{0.72*10^{-4}}[/tex]

           [tex]P = 583333 \ N/m^2[/tex]

Answer:

The observer could see to a depth of 4.38 m

Explanation:

Please check attachment for diagram.

Mathematically, from Snell law;

n1sin theta = n2 sin theta

1 * sin 90 = n2 * sin θR

where n2 = 1.33

1/1.33 = sin θR

Sin θR = 0.7519

θR = arc sin 0.7519

θR = 48.76

Now to get the height, we use the triangle

Using trigonometric ratio;

Tan( 90- θR) = H/5

H = 5 Tan( 90 - θR)

H = 5 Tan( 90-48.76)

H = 5 Tan41.24

H = 4.38 m

Answer:

L = 0.8 m

Explanation:

Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:

Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m

but,

Δx = λL/d

λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m

d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m

L = Distance between slits and screen = ?

Therefore, using the values, we get:

2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)

L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)

L = 0.8 m

In a closed system the mass is conserved, but the energy is not conserved.

To find the answer, we have to study about different systems in thermodynamics.

Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.

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Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

Answer:

a. 0.058°

b.  0.117°

Explanation:

a. The angular position of the first-order is:

[tex] d*sin(\theta) = m\lambda [/tex]

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]

Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.

b. The angular position of the second-order is:

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]

Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.

I hope it helps you!

Answer:

D Is 430m

Explanation:

See attached file

Answer:

The smallest value is n= 2

Explanation:

The balmer equation is given below

1/λ = R(1/4 - 1/n₂²).

R= 1.0973731568508 × 10^7 m^-1

λ= 400*10^-9 m

(400*10^-9)= 1.0973731568508 × 10^7 (1/4-1/n²)

(400*10^-9)/1.0973731568508 × 10^7

= 1/4 - 1/n²

364.51 *10^-16= 1/4 - 1/n²

1/n²= 1/4 -364.51 *10^-16

1/n² = 0.25-3.6451*10^-14

1/0.25= n²

4= n²

√4= n

2= n

The smallest value is N= 2

Answer:

The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

Explanation:

To find the speed of light in air and in polystyrene we need to use the following equation:

[tex] c_{m} = \frac{c}{n} [/tex]

Where:

[tex]c_{m}[/tex]: is the speed of light in the medium

n: is the refractive index of the medium

In air:

[tex]c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s[/tex]

In polystyrene:

[tex]c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s[/tex]  

Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

I hope it helps you!

Answer

82.3 wavelengths

Answer:

Explanation:

The moment of inertia of the disk I  = 1/2 m R² where R is radius of the disc and m is its mass .

putting the values

I = .5 x 95 x .38²

= 6.86 kg m²

n = 87 rpm = 87 / 60 rps

n = 1.45 rps

angular velocity ω = 2π n , n is frequency of rotation .

= 2 x 3.14 x 1.45

= 9.106 radian /s

frictional force = 16 x .2

= 3.2 N

torque created by frictional force = 3.2 x .38

= 1.216 N.m

angular acceleration = torque / moment of inertia

= - 3.2 / 6.86

α  = - 0.4665 rad /s²

b ) ω² = ω₀² +  2 α θ , where α is angular acceleration

0 = 9.106² - 2 x .4665 θ

θ = 88.87 radian

no of turns = 88.87 / 2π

= 14.15  turns

Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

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The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

                      [tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]

Substitute,  [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]

                   [tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]

Hence, The new length of aluminum rod is 17.435 cm.

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https://brainly.com/question/19495810

Answer:

[tex]\huge\boxed{Refracted}[/tex]

Explanation:

When a ray of light passes from glass to water, it

1) is Slightly refracted (bending of light)

2) moves away from the normal.

Whenever a light ray travels from a denser medium to a rarer medium, it bends away from the normal.

Answer:

refraction

Explanation: