A new experiment is created with the screen at a distance of 1.9 m from the slits (with spacing 0.11 mm). What is the distance between the second order bright fringe of light with λ = 695 nm and the third order bright fringe of light with λ = 407 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.) |x| =

Answer:

24.616 mm

12.384 mm

Explanation:

[tex]D_b[/tex] = Blanking die diameter = 25 mm

[tex]D_h[/tex] = Punch diameter = 12 mm

a = Allowance = 0.06 for aluminium

t = Thickness = 3.2 mm

Clearance

[tex]c=at\\\Rightarrow c=0.06\times 3.2\\\Rightarrow c=0.192[/tex]

Diameter of punch is

[tex]d_p=D_b-2c\\\Rightarrow d_p=25-2\times 0.192\\\Rightarrow d_p=24.616\ mm[/tex]

Diameter of blanking punch is 24.616 m

For punching operation

[tex]d_d=D_h+2c\\\Rightarrow d_d=12+2\times 0.192\\\Rightarrow d_d=12.384\ mm[/tex]

The diameter of punching die is 12.384 mm

For the blanking operation, the punch diameter should be 12.0 mm and the die diameter should be 25.0 mm. For the punching operation, the punch diameter should be 12.0 mm and the die diameter should be slightly larger than 12.0 mm, typically by a clearance of 0.05 mm to 0.15 mm per side.

(a) Blanking Operation:

The blanking operation involves cutting the outside profile of the washer from the aluminum sheet stock. The punch and die sizes for this operation are determined by the outside and inside dimensions of the washer.

- The die diameter for the blanking operation should match the outside diameter of the washer, which is 25.0 mm. This is the size of the hole in the die through which the punch will pass to cut the washer's outer edge.

- The punch diameter for the blanking operation should match the inside diameter of the washer, which is 12.0 mm. This is the size of the punch that will push through the sheet stock to create the inner hole of the washer.

(b) Punching Operation:

The punching operation involves cutting the inside hole of the washer after the blanking operation has been completed. The punch and die sizes for this operation are determined by the inside diameter of the washer and the required clearance.

 - The punch diameter for the punching operation should also be 12.0 mm, the same as the inside diameter of the washer. This ensures that the punch cuts the material exactly at the desired inner diameter.

- The die diameter for the punching operation should be slightly larger than the punch diameter to allow for clearance. Clearance is necessary to prevent the punch and die from binding together and to allow for the removal of the slug (the piece of material that is punched out). The typical clearance ranges from 0.05 mm to 0.15 mm per side. Therefore, the die diameter would be 12.0 mm plus two times the clearance. If we take the middle value of the clearance range, which is 0.10 mm per side, the die diameter would be 12.0 mm + 2 * 0.10 mm = 12.20 mm.

Answer:

I₂ = 25.4 W

Explanation:

Polarization problems can be solved with the malus law

     I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

    θ₂ = 68.0-22.2 = 45.8º

    I = I₂ cos² θ₂

    I₂ = I / cos₂ θ₂

    I₂ = 75.5 / cos² 45.8

    I₂ = 155.3 W

We repeat for First and second polarizer

   I₂ = I₁ cos² θ₁

   I₁ = I₂ / cos² θ₁

   I₁ = 155.3 / cos² 22.2

   I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

    I₁ = I₀ / 2

   I₀ = 2 I₁

   I₀ = 2 181.2

   I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

    I₁ = 181.2 W

The intensity at the exit is

    I₂ = I₁ cos² θ₂

    I₂ = 181.2 cos² 68.0

   I₂ = 25.4 W

Explanation:

It is given that,

Force exerted by a boy, F = 11.8 N

Angle above the horizontal, [tex]\theta=28^{\circ}[/tex]

Mass of the sled, m = 6.15 kg

Distance moved, d = 2.75 m

Initial speed, u = 0.37 m/s

Let W is the work done by the boy. Using the expression for the work done to find it as :

[tex]W=Fd\ cos\theta[/tex]

[tex]W=11.8\times 2.75\ cos(28)[/tex]

W = 28.65 joules

Let v is the final speed of the sled. Using the work energy theorem to find it. It states that the work done is equal to the change in kinetic energy of an object. It is given by :

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]v=\sqrt{\dfrac{2W}{m}+u^2}[/tex]

[tex]v=\sqrt{\dfrac{2\times 28.65}{6.15}+(0.37)^2}[/tex]

v = 3.07 m/s

So, the final speed of the sled after it moves 2.75 m is 3.07 m/s. Hence, this is the required solution.

The final speed of the sled is approximately 3.073 m/s and the the work done by the boy on the sled is 28.64 J.

To solve this problem, we need to calculate the work done by the boy on the sled and then use that to find the final speed of the sled.

First, let's calculate the work done by the boy. Work (W) is defined as the product of the force (F) applied in the direction of motion and the distance (d) over which the force is applied. The force applied in the direction of motion is the horizontal component of the force exerted by the boy. We can calculate this using trigonometry:

[tex]\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \][/tex]

[tex]\[ F_{\text{horizontal}} = 11.8 \, \text{N} \cdot \cos(28.0^\circ) \][/tex]

[tex]\[ F_{\text{horizontal}} \approx 11.8 \, \text{N} \cdot 0.8829 \][/tex]

[tex]\[ F_{\text{horizontal}} \approx 10.42 \, \text{N} \][/tex]

Now, we can calculate the work done:

[tex]\[ W = F_{\text{horizontal}} \cdot d \][/tex]

[tex]\[ W = 10.42 \, \text{N} \cdot 2.75 \, \text{m} \][/tex]

\[ W \approx 28.64 \, \text{J} \]

Next, we need to find the final speed of the sled. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (KE):

[tex]\[ W = \Delta KE \][/tex]

[tex]\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \][/tex]

Given that the initial speed [tex]\( v_i \)[/tex] is 0.370 m/s and the mass [tex]\( m \)[/tex] of the sled is 6.15 kg, we can solve for the final speed [tex]\( v_f \)[/tex]:

[tex]\[ 28.64 \, \text{J} = \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot v_f^2 - \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot (0.370 \, \text{m/s})^2 \][/tex]

[tex]\[ 28.64 \, \text{J} = 3.075 \, \text{kg} \cdot v_f^2 - 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]

[tex]\[ 28.64 \, \text{J} + 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]

[tex]\[ 29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]

[tex]\[ v_f^2 = \frac{29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{3.075 \, \text{kg}} \][/tex]

[tex]\[ v_f^2 \approx 9.444 \, \text{m}^2/\text{s}^2 \][/tex]

[tex]\[ v_f \approx \sqrt{9.444 \, \text{m}^2/\text{s}^2} \][/tex]

[tex]\[ v_f \approx 3.073 \, \text{m/s} \][/tex]

Answer

given,

length of wire = 10 cm = 0.1 m

resistance of the wire =  0.330 ohms

speed of pulling = 4 m/s

Power = 4.40 W

a) Force of pull = ?

 [tex]P = F_{pull} v[/tex]

 [tex]F_{pull} =\dfrac{P}{v}[/tex]

 [tex]F_{pull} =\dfrac{4.40}{4}[/tex]

 [tex]F_{pull} =1.1\ N[/tex]

b) using formula

  [tex]P = \dfrac{B^2l^2v^2}{R}[/tex]

  where B is the magnetic field

   v is the pulling velocity

   R is the resistance of the wire

  [tex]B =\sqrt{\dfrac{PR}{l^2v^2}}[/tex]

  [tex]B =\sqrt{\dfrac{F_{pull} \times v R}{l^2v^2}}[/tex]

  [tex]B =\sqrt{\dfrac{F_{pull} \times R}{l^2v}}[/tex]

  [tex]B =\sqrt{\dfrac{1.1 \times 0.33}{0.1^2\times 4}}[/tex]

  [tex]B =\sqrt{9.075}[/tex]

      B = 3.01 T

The pulling force in this magnetic field is equal to 1.1 Newton.

Given the following data:

Mathematically, the pulling force in a magnetic field is given by this formula:

[tex]F = \frac{Power}{Speed} \\\\F=\frac{4.40}{4.00}[/tex]

F = 1.1 Newton.

To determine the strength of the magnetic field, we would apply this formula:

[tex]B=\sqrt{\frac{PR}{L^2V^2} }\\\\B=\sqrt{\frac{FVR}{L^2V^2} }\\\\B=\sqrt{\frac{FR}{L^2V} }[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]B=\sqrt{\frac{1.1 \times 0.330}{0.1^2 \times 4.00} } \\\\B=\sqrt{\frac{0.363}{0.01 \times 4.00} }\\\\B=\sqrt{9.075}[/tex]

B = 3.012 T.

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Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

[tex]f_o[/tex] = Focal length of objective = 0.9 cm

[tex]f_e[/tex] = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

[tex]m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52[/tex]

The angular magnification of the compound microscope is -252.52

The value of all options are mathematically given as

a) a=1.5 rad/s^2

b)theta=248 rad

c) theta '=248 radians

d) angles in degree=14207.511

Question Parameter(s):

Generally, the equation for the final angular speed  is mathematically given as

w2=w+at

Therefore

37=25+a 8

a=1.5 rad/s^2

b)

Average angular speed

[tex]\theta =\omega _1t+\frac{1}{2}\alpha t^2[/tex]

[tex]\theta =25\times 8+\frac{1}{2}\times 1.5\times 8^2[/tex]

theta=248 rad

Hence

theta =200+48

theta=248 rad

In conclusion,

c) theta '=248 radians

d) angles in degree=14207.511

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Final answer:

The angular acceleration is 1.5 rad/s², the average angular speed is 31 rad/s, the angle through which the wheel rotates is 320 rad, and the angle in degrees is 18333 degrees.

Explanation:

(a) To find the angular acceleration, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time interval. Plugging in the given values, we get: angular acceleration = (37 rad/s - 25 rad/s) / 8 s = 1.5 rad/s².

(b) The average angular speed can be found by taking the average of the initial and final angular velocities. So, the average angular speed = (25 rad/s + 37 rad/s) / 2 = 31 rad/s.

(c) The angle through which the wheel rotates can be found using the formula: angle = initial angular velocity * time + (1/2) * angular acceleration * time². Plugging in the given values, we get: angle = 25 rad/s * 8 s + (1/2) * 1.5 rad/s² * (8 s)² = 320 rad.

(d) To convert radians to degrees, we use the conversion factor: 1 radian = 180 degrees / pi. So, the angle in degrees = 320 rad * 180 degrees / pi = 18333 degrees (rounded to the nearest whole number).

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m[/tex]

The image distance is 0.051 m

When u = 50 cm

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m[/tex]

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

To focus on the closer object, the lens must move away from the object. By applying the lens formula, we calculate that the lens must move exactly 7mm away from the initial position.

This problem comes down to the concept of the lens formula in optical physics. The lens formula is 1/f = 1/v - 1/u where f is the focal length, v is the image distance, and u is the object distance.

Part A: To focus on the second object which is closer, the lens must move away from the object.

Part B: To calculate how far the lens must move, we need to apply the lens formula. Let's calculate the image distances for the two object distances.

For an object at 2.5 m, 1/v = 1/f + 1/u = 1/0.05 + 1/2.5 = 20 + 0.4 = 20.4. So, v = 0.049 m or 49 mm.

For an object at 0.5 m, 1/v = 1/f + 1/u = 1/0.05 - 1/0.5 = 20 - 2 = 18. So, v = 0.056 m or 56 mm.

Hence, the lens must move 56mm - 49mm = 7mm away from the initial position.

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Answer:

hi sandra!!

Explanation:

the number 3 is the correct!!

The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earth’s atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.

Auroras that occur in the northern hemisphere are called ‘Aurora Borealis’ or ‘northern lights’ and auroras that occur in the southern hemisphere are called ‘Aurora Australis’ or ‘southern lights’.

Auroral displays can appear in many differents colours, but green is the most common. Colours such as red, yellow, green, blue and violet are also seen occasionally. The auroras can appear in many forms, from small patches of light that appear out of nowhere to streamers, arcs, rippling curtains or shooting rays that light up the sky with an incredible glow.

Answer:

Option-(3): The aurora borealis are the formation of the colored patterns on the pole of the planet, but other celestial bodies also experience such effects due to the heat flares or the sun flares when exposed to them.

Explanation:

Aurora borealis:

The natural phenomenon which is caused in most of the celestial bodies, but when our planet earth is exposed to such electromagnetic rays or flares released from the Sun, when the energy level on the surface increases and the flare is cut off from the environment of the sun. As, starts motion towards the different celestial bodies present in the solar system.

And when it reaches earth, the earth surface reflects the high in energy atoms coming towards its atmosphere. As the magnetic field present on the earth surface interacts with the Sun's flare and then it cut off's into two moving towards the two poles of the planet.

To find the angular acceleration of the cylinder, calculate the moment of inertia of the cylinder, determine the torque due to the falling bucket, and then use Newton's second law for rotation to find the angular acceleration.

The subject involves calculating the angular acceleration of a cylinder when the bucket of water tied to a rope around the cylinder is let go. We assume no friction or air resistance in this case.

First, we need to calculate the moment of inertia (I) of the solid cylinder, which can be calculated using the formula I = 0.5 * m * r^2. Here, m is the mass of the cylinder (50 kg) and r is the radius (half of the diameter, 0.125 m). After calculating I, use Newton's second law for rotation τ = I * α, where τ is torque and α is the angular acceleration.

The torque τ can be calculated as the product of the force exerted by the falling bucket (F) and the radius r of the cylinder. The force F is the weight of the falling bucket, m * g, where m is the mass of the bucket (20 kg) and g is acceleration due to gravity (about 9.8 m/s^2). Once you have calculated τ, you can solve for α by rearranging the equation to α = τ/I.

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The angular acceleration of the 50-kg cylinder is found to be approximately 62.7 rad/s².

To find the angular acceleration of the 50-kg cylinder when the bucket of water falls down to the bottom of the well, we will use the concept of rotational dynamics. The moment of inertia (I) of a solid cylinder is given by:

I = (1/2) [tex]\times[/tex] M[tex]\times[/tex] R²

where M is the mass and R is the radius. Given the mass M is 50 kg and the diameter is 25 cm, so the radius R is 0.125 m:

I = (1/2) * 50 kg[tex]\times[/tex] (0.125 m)² = 0.390625 kg m²

The torque (τ) caused by the falling bucket is:

The force exerted by the bucket of water is its weight, F = m [tex]\times[/tex] g, where m is the mass of the bucket and water (20 kg) and g is the acceleration due to gravity (9.8 m/s²):

So,

Using Newton's second law for rotation:

where α is the angular acceleration:

Solving for α:

Therefore, the angular acceleration of the 50-kg cylinder is approximately 62.7 rad/s².

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

Answer:

b = 1.2 m

L = 4.8 m

d= 1.2 m

Explanation:

Lets take

Width of the tank = b

Depth = d

length = L

Given that L = 4 b

v= 2 cm/s

Flow rate Q= 10,000 m³/d

We know that 1 d = 24 hr = 24 x 3600 s

1 d= 86400 s

Q= 0.115  m³/s

Flow rate Q

Q= Area x velocity

Q= L . b .v

0.115 = 4 b . b . 0.02

4 b² = 5.78  

b = 1.2 m

L = 4.8 m

d= 1.2 m

This is the dimension of the tank.

To design the settling tank with the given water flow and settling velocity, we use conservation of mass and settling speed to calculate the dimensions. We convert the flow rate to m3/s, relate it to the tank's cross-sectional area, and given settling characteristics to find the tank's width, length, and depth.

To design a rectangular settling tank for settling sand particles with a settling velocity of 2.0 cm/s and a water flow of 10,000 m3/d, we will apply the principles of conservation of mass and the given settling velocity to determine the dimensions of the tank. The problem states that the tank length (L) should be four times the width (W), and the width should be equal to the depth (D), therefore L = 4W and W = D.

To find the dimensions, we first convert the water flow rate into m3/s: flow rate = 10,000 m3/d × (1 day/86400 s) ≈ 0.1157 m3/s. Since the sand particles settle at 2 cm/s, for a particle to settle before reaching the end of the tank, the residence time should be at least the depth divided by the settling velocity. From here, we use the flow rate (Q), which is also equal to the cross-sectional area (A) times the water velocity (V), and we know that A = W² for this tank because W = D.

Therefore, Q = W²V. Using V = L / (residence time) and substituting for L = 4W, we find that Q = W² × (4W / (W/0.02 m/s)) which simplifies to Q = 0.08 m/s × W³. Solving for W gives us W = ∛(Q / 0.08 m/s) and subsequently we find L = 4W and D = W. Finally, we can plug in the flow rate and solve for the dimensions in meters.

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Answer:

The force the bat applies to the ball is 340 N

Explanation:

given information:

mass of the ball, m = 400 g = 0.4 kg

initial velocity, [tex]v_{i}[/tex] = 30 m/s

final velocity, [tex]v_{f}[/tex]  = 38 m/s

time, t = 80 ms = 0.08 s

the correlation between the change in momentum and the force is shown by the following equation

ΔP = FΔt

F = ΔP/Δt

where,

Δp = mΔv

     = m([tex]v_{f} - v_{i}[/tex])

     = m([tex]v_{f} + v_{i}[/tex])

[tex]v_{i}[/tex] = - because in opposite direction, thus

F = ΔP/Δt

  = m([tex]v_{f} + v_{i}[/tex])/Δt

   = (0.4)(38+30)/0.08

   = 340 N

Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles

Explanation:

Answer:209.98 kJ

Explanation:

mass of water [tex]m=456 gm[/tex]

Initial Temperature of Water [tex]T_i=25^{\circ}C[/tex]

Final Temperature of water [tex]T_f=-10^{\circ}C[/tex]

specific heat of ice [tex]c=2090 J/kg-K[/tex]

Latent heat [tex]L=33.5\times 10^4 J/kg[/tex]

specific heat of water [tex]c_{water}=4.184 KJ/kg-K[/tex]

Heat require to convert water at [tex]T=25^{\circ}C[/tex] to [tex]T=0^{\circ}C[/tex]

[tex]Q_1=0.456\times 4.184\times (25-0)=47.69 kJ[/tex]

Heat require to convert water at [tex]T=0^{\circ}[/tex] to ice at [tex]T=0^{\circ}[/tex]

[tex]Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ[/tex]

heat require to convert ice at [tex]T=0^{\circ} C\ to\ T=-10^{\circ} C[/tex]

[tex]Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ[/tex]

Total heat [tex]Q=Q_1+Q_2+Q_3[/tex]

[tex]Q=47.69+152.76+9.53=209.98 kJ[/tex]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity  = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2[/tex]

v = r ω

[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}[/tex]

[tex]I = \dfrac{m(2gh - v^2)r^2}{v^2}[/tex]

[tex]I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}[/tex]

[tex]I =mr^2(0.287)[/tex]

I = 0.287 MR²

Answer:

I= 6.292  kg.m²

Explanation:

Given that

m = 1.3 kg

Side of square a= 1.1 m

The distance r

[tex]r=\sqrt{{a^2}+{a^2}}[/tex]

[tex]r={a}{\sqrt 2}[/tex]

[tex]r={1.1}{\sqrt 2}[/tex]

The moment of inertia I

The axis passes through one of the mass then the distance of the that mass from the axis will be zero.

I = m a² + m a² + m r²

By putting the values

I = m a² + m a² + m r²

I =m( 2 a² +  r²)

I =1.3( 2 x 1.1² +  2 x 1.1²)

I = 1.3 x 4  x 1.1² kg.m²

I= 6.292  kg.m²

Answer:

108507.02596 Pa

42.42 cm

Explanation:

F = Force

m = Mass of piston = 23 kg

[tex]T_1[/tex] = Initial temperature = 307°C

[tex]T_2[/tex] = Final temperature = 13°C

[tex]P_o[/tex] = Outside pressure

[tex]P_i[/tex] = Pressure inside cylinder

A = Area of pistion

h = Height of piston

Change in pressure is given by

[tex]\Delta P=\frac{F}{A}\\\Rightarrow \Delta P=\frac{mg}{\pi r^2}\\\Rightarrow \Delta P=\frac{23\times 9.81}{\pi 0.1^2}\\\Rightarrow \Delta P=7182.02596\ Pa[/tex]

[tex]P_i-P_o=\Delta P\\\Rightarrow P_i=\Delta P+P_0\\\Rightarrow P_i=7182.02596+101325\\\Rightarrow P_i=108507.02596\ Pa[/tex]

Gas pressure inside the cylinder is 108507.02596 Pa

From ideal gas law we have the relation

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow \frac{Ah_1}{T_1}=\frac{Ah_2}{T_2}\\\Rightarrow \frac{h_1}{T_1}=\frac{h_2}{T_2}\\\Rightarrow h_2=\frac{h_1}{T_1}\times T_2\\\Rightarrow h_2=\frac{0.86}{307+273.15}\times (13+273.15)\\\Rightarrow h_2=0.42418\ m=42.42\ cm[/tex]

The height of the piston at 13°C will by 42.42 cm

(a) The gauge pressure inside the cylinder is 108,500 Pa.

(b) The height of the piston when the temperature is lowered to 13°C is 42.4 cm.

The change in the pressure of the gas inside the cylinder is calculated as follows;

ΔP = F/A

ΔP = mg/πr²

ΔP = (23 x 9.8)/(π x0.1²)

ΔP = 7,174.7 Pa

The gauge pressure inside the cylinder is calculated as follows;

Pi = ΔP + Po

Pi = 7,174.7  +  101325

Pi = 108,500 Pa

The height of the piston is calculated as follows;

[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2} \\\\\frac{Ah_1}{T_1} = \frac{Ah_2}{T_2} \\\\h_2 = \frac{T_2h_1}{T_1} \\\\h_2 = \frac{(13 + 273) \times 0.86}{(307 + 273)} \\\\h_2 = 0.424 \ m[/tex]

h₂ = 42.4 cm

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Answer:

A,

Explanation:

To solve the exercise it is necessary to use the concepts and definitions given for electromagnetic energy, in which

[tex]E = \frac{hc}{\lambda}[/tex]

Where,

h = Plank's constant [tex](6.626*10^{-34}J.s)[/tex]

c = Speed of light [tex](3*10^8m/s)[/tex]

[tex]\lambda[/tex] =Wavelength

If we analyze these characteristics both h and c are constant so the energy is inversely proportional to the size of the wave.

The larger the amplitude of the wave, the smaller the energy.

On the other hand we have the frequency value defined as

[tex]f = \frac{c}{\lambda}[/tex]

In this case the frequency is also inversely proportional to the wavelength.

In this case, the amplitude of the largest wave is infrared, so it will have less energy and less frequency. The fact that it has a low frequency by the wavelength, also generates that it has a low energy. But not because it has a large wavelength, on the contrary, because its wavelength is small.

In the case of the ultraviolet wave it will have greater frequency and greater energy. Therefore of all the options, A is the only one valid.

Answer:

138.18 minutes

Explanation:

[tex]L_v[/tex] = Latent heat of water at 0°C = 80 cal/g

m = Mass of water = 570 g

Heat removed for freezing

[tex]Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal[/tex]

Let N be the number of cycles and each cycle removes 56 cal from the freezer.

So,

[tex]55\times N=45600\\\Rightarrow N=\frac{45600}{55}[/tex]

Each cycle takes 10 seconds so the total time would be

[tex]\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes[/tex]

The total time taken to freeze 138.18 minutes

Answer:

heat absorbed by water = 2.39 KJ

Explanation:

given,                                    

mass of molten NaCl = 5 g

mass of  water = 25 g                    

Temperature of water = 25.0 °C

Temperature increased to = 47.9 °C

heat absorbed by water = m S ΔT

 S is specific heat in J/gm °C                

 Specific heat of water = 4.184 J/gm °C

 m is mass of water                                            

heat absorbed by water = 25 x 4.184 x (47.9-25)

                                         = 25 x 4.184 x 22.9

                                         = 2395.34 J

heat absorbed by water = 2.39 KJ

The amount of heat transferred is 2.39 KJ.

This is the form of energy which is transferred from one body to another as the result of a difference in temperature.

mass of molten NaCl = 5 g

mass of  water = 25 g                    

Temperature of water = 25.0 °C

Final Temperature of water = 47.9 °C

Specific heat of water = 4.184 J/gm °C

Heat = m S ΔT

Heat absorbed by water = 25 x 4.184 x (47.9 -25)

= 25 x 4.184 x 22.9 = 2395.34 J

= 2.39 KJ.

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Answer:3.51

Explanation:

Given

Coefficient of Friction [tex]\mu =0.4 [/tex]

Consider a small element at an angle \theta having an angle of [tex]d\theta [/tex]

Normal Force[tex]=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}[/tex]

[tex]N=T\cdot d\theta [/tex]

Friction [tex]f=\mu \times Normal\ Reaction[/tex]

[tex]f=\mu \cdot N[/tex]

and [tex]T+dT-T=f=\mu Td\theta [/tex]

[tex]dT=\mu Td\theta [/tex]

[tex]\frac{dT}{T}=\mu d\theta [/tex]

[tex]\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta [/tex]

[tex]\frac{T_2}{T_1}=e^{\mu \pi}[/tex]

[tex]\frac{T_2}{T_1}=e^{0.4\times \pi }[/tex]

[tex]\frac{T_2}{T_1}==e^{1.256}[/tex]

[tex]\frac{T_2}{T_1}=3.51[/tex]

Answer:L=109.16 m

Explanation:

Given

initial temperature [tex]=20^{\circ}C[/tex]

Final Temperature [tex]=80^{\circ}C[/tex]

mass flow rate of cold fluid [tex]\dot{m_c}=1.2 kg/s[/tex]

Initial Geothermal water temperature [tex]T_h_i=160^{\circ}C[/tex]

Let final Temperature be T

mass flow rate of geothermal water [tex]\dot{m_h}=2 kg/s[/tex]

diameter of inner wall [tex]d_i=1.5 cm[/tex]

[tex]U_{overall}=640 W/m^2K[/tex]

specific heat of water [tex]c=4.18 kJ/kg-K[/tex]

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

[tex]\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)[/tex]

[tex]2\times (160-T)=1.2\times (80-20)[/tex]

[tex]160-T=36[/tex]

[tex]T=124^{\circ}C[/tex]

As heat exchanger is counter flow therefore

[tex]\Delta T_1=160-80=80^{\circ}C[/tex]

[tex]\Delta T_2=124-20=104^{\circ}C[/tex]

[tex]LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}[/tex]

[tex]LMTD=\frac{80-104}{\ln \frac{80}{104}}[/tex]

[tex]LMTD=91.49^{\circ}C[/tex]

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

[tex]\dot{m_c}c(80-20)=U\cdot A\cdot[/tex] (LMTD)

[tex]A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2[/tex]

[tex]A=\pi DL=5.144[/tex]

[tex]L=\frac{5.144}{\pi \times 0.015}[/tex]

[tex]L=109.16 m[/tex]

The length of a counterflow double-pipe heat exchanger required can be calculated by first determining the heat needed to raise the water temperature, then calculating the necessary surface area, and finally dividing by the inner circumference of the pipe to find the length.

To determine the length of the heat exchanger required to heat water from 20°C to 80°C using geothermal water at 160°C, we can apply the basic principles from thermodynamics and heat transfer. The equation that relates the heat transfer Q with the overall heat transfer coefficient U, the surface area A, and the temperature difference ΔT is Q = U * A * ΔT. We can determine the heat needed to raise the temperature of water by using the specific heat of water and the mass flow rate of water.

First, the amount of heat needed to raise the temperature of water (Q) can be calculated using the specific heat capacity of water (c_p, which is approximately 4.18 kJ/kgK), the mass flow rate of water (m_dot), and the temperature rise (ΔT = T_f - T_i), using the formula Q = m_dot * c_p * ΔT.

Next, we can rearrange the heat transfer equation to solve for the surface area A required for the heat exchanger: A = Q / (U * ΔTLMTD), where ΔTLMTD is the logarithmic mean temperature difference, which can be calculated for a counterflow heat exchanger knowing the inlet and outlet temperatures of both the hot and cold fluids.

Finally, we can find the length of the heat exchanger (L) by dividing the surface area A by the inner circumference of the pipe (π * d), with d being the inner pipe diameter.

By calculating the heat necessary for the temperature rise of the water and applying the given overall heat transfer coefficient, we can determine the required length of the exchanger.

Answer:

λ = 547.96 nm

Explanation:

given,

thickness of soap bubble = 103 nm

refractive index of thin film = 1.33

using formula of constructive interference

[tex]2 n t = (m + \dfrac{1}{2})\times \lambda[/tex]

t is thickness of the medium

n is refractive index

m = 0,1,2...

now,

[tex]2 n t = (m + 0.5)\times \lambda[/tex]

at m = 0

[tex]2\times 1.33 \times 103 =0.5\times \lambda[/tex]

λ = 547.96 nm

at m = 1

[tex]2 n t = (m + 0.5)\times \lambda[/tex]

[tex]2\times 1.33 \times 103 =1.5\times \lambda[/tex]

λ = 182.65 nm

the only visible light enhanced by the thin film is of wavelength

λ = 547.96 nm

Answer:

The resistance must be 6.67[tex]\Omega[/tex]

Solution:

Resistance, [tex]R_{1} = 20\Omega[/tex]

Resistance, [tex]R_{2} = 10\Omega[/tex]

For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:

Thus in parallel:

[tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]

[tex]\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}[/tex]

[tex]R_{eq} = 6.67\ \Omega[/tex]

Answer:

68.6 m/s

Explanation:

v = Speed of sound in air = 343 m/s

u = Speed of train

[tex]f_1[/tex] = Actual frequency

From the Doppler effect we have the observed frequency as

When the train is approaching

[tex]f=f_1\frac{v+u}{v}[/tex]

When the train is receeding

[tex]\frac{2f}{3}=f_1\frac{v-u}{v}[/tex]

Dividing the above equations we have

[tex]\frac{f}{\frac{2f}{3}}=\frac{f_1\frac{v+u}{v}}{f_1\frac{v-u}{v}}\\\Rightarrow \frac{3}{2}=\frac{v+u}{v-u}\\\Rightarrow 3v-3u=2v+2u\\\Rightarrow v=5u\\\Rightarrow u=\frac{v}{5}\\\Rightarrow u=\frac{343}{5}\\\Rightarrow u=68.6\ m/s[/tex]

The speed of the train is 68.6 m/s

Answer

given,

Distance for decibel reading

 r₁ = 13 m

 r₂ = 24 m

When the engineer is at r₁ reading is β₁ = 101 dB

now, Calculating the Intensity at r₁

Using formula

[tex]\beta = 10 log(\dfrac{I}{I_0})[/tex]

I₀ = 10⁻¹² W/m²

now inserting the given values

[tex]101= 10 log(\dfrac{I}{10^{-12}})[/tex]

[tex]10.1= log(\dfrac{I}{10^{-12}})[/tex]

[tex]\dfrac{I}{10^{-12}}= 10^{10.1}[/tex]

[tex]I =10^{10.1} \times 10^{-12}[/tex]

I = 0.01258 W/m²

now, calculating power at r₁

P₁ = I₁ A₁

P₁ = 0.01258 x 4 π r²

P₁ = 0.01258 x 4 π x 13²

P₁ = 26.72 W

Final answer:

The question deals with acoustics in physics, where an audio engineer uses the decibel scale to measure sound pressure levels and requires an understanding of the inverse square law and logarithmic measurements of sound intensity.

Explanation:

The student's question involves the concept of sound intensity levels and decibels, which are part of the field of acoustics, a topic in physics. Specifically, the student is dealing with the measurement of sound pressure levels at various distances from a sound source, which is an application of the inverse square law in acoustics. The decibel (dB) scale is a logarithmic unit used to measure sound intensity, with 0 dB representing the threshold of human hearing at an intensity of 10-12 W/m².

The loudness experienced by the audio engineer at various distances can be computed using the formula for sound intensity level (SIL), which is SIL = 10 log(I/I0), where I is the sound intensity and I0 is the reference intensity of 10-12 W/m².

Answer:

[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]

Explanation:

The mass of object is m

The coefficient of friction is μ

the speed of object at x= 0 is v

when the object compress the springs stops so,

The kinetic energy of the spring is equal to and friction work is done

[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2+\mu(mg)x[/tex]...........

after that it returns from the rest and finally comes to rest after coming to x=0

position.

So, now the potential energy of the spring is equal to the work done by the friction force.

[tex]\frac{1}{2}kx^2= \mu mgx[/tex]

[tex]\frac{1}{2}kx= \mu mg[/tex]

[tex]x= \frac{2\mu mg}{x}[/tex]

substitute the x value in equation 1

mv^2 = kx^2+2\mu (mg)x

[tex]mv^2= k\frac{2\mu mg}{k}+ 2\mu mg \frac{2\mu mg }{k}[/tex]

solving this we get

[tex]mv^2= \frac{8m^2\mu^2g^2}{k}[/tex]

therefore the force constant

[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]

Spring Constant is the measure of a spring's stiffness. The value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].

Spring Constant is the measure of a spring's stiffness.

Given to us

Mass of object = m

The coefficient of friction = μ

The speed of the object at (x= 0) = v

When the object compresses the spring after a distance the spring and the object both stop, therefore, the Kinetic energy of the object transforms to the kinetic energy of the spring during compression and the potential energy when fully compressed, therefore,

The kinetic energy of the object during the compression,

[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x[/tex]

The kinetic energy of the object after complete compression,

[tex]\dfrac{1}{2}kx^2 = \mu m g x\\\\\dfrac{1}{2}kx = \mu m g \\\\x = \dfrac{2 \mu m g}{k}[/tex]

Substitute the value of x in the kinetic energy of the object during the compression,

[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x\\\\\dfrac{1}{2} mv^2 = \dfrac{1}{2}k(\dfrac{2 \mu m g}{k})^2 + \mu (mg)(\dfrac{2 \mu m g}{k})\\\\\\mv^2 = \dfrac{8m^2 \mu ^2 g^2}{k}\\\\k= \dfrac{8m^2 \mu ^2 g^2}{mv^2 }\\\\k= \dfrac{8m \mu ^2 g^2}{v^2 }[/tex]

Hence, the value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].

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Final answer:

In Situation B, the speed of the block at x = x0e is 3.5 m/s.

Explanation:

In Situation A, a 20-kg block is pressed against a single spring and released. The speed of the block at x = x0e is va,f = 5.0 m/s. In Situation B, the same 20-kg block is pressed against two identical springs in series. The entire spring assembly is compressed by Δx = 10 cm. The question asks for the speed of the block at x = x0e, which we can find using the principle of conservation of mechanical energy.

The speed of the block in Situation B, vb,f, can be found as follows:

Solving this equation gives us vb,f = 3.5 m/s.

Answer:

It slows down from 0-31.25s

It speeds up from 31.25-62.507s

Explanation:

If we find the maximum of the equation, we will know the moment when it changes direction. It will slow down on the first interval, and speed up on the second one.

[tex]Y = -16t^2+1000t+7[/tex]

[tex]Y' = -32t+1000 = 0[/tex]  Solving for t:

t = 31.25s

The bullet will slow down in the interval 0-31.25s

Let's now find the moment when it hits ground:

[tex]Y = 0 = -16t^2+1000t+7[/tex]   Solving for t:

t1 = -0.0069s     t2 = 62.507s

The bullet will speed up in the interval 31.25-62.507s

Answer:

[tex]1\times 10^{8}\ Pa[/tex]

Explanation:

k = Boltzmann constant = [tex]1.38\times 10^{-23}\ J/K[/tex]

r = Radius of gas molecule = [tex]2\times 10^{-10}\ m[/tex]

t = Temperature = 300 K

P = Pressure

Volume of gas per molecule is given by

[tex]V=\frac{4}{3}\pi r^3\\\Rightarrow V=\frac{4}{3}\pi (2\times 10^{-10})^3\\\Rightarrow V=3.35103\times 10^{-29}\ m^3[/tex]

From the ideal gas law we have

[tex]PV=kt\\\Rightarrow P=\frac{kt}{V}\\\Rightarrow P=\frac{1.38\times 10^{-23}\times 300}{3.35103\times 10^{-29}}\\\Rightarrow P=123544104.35\ Pa=1\times 10^{8}\ Pa[/tex]

The pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior is [tex]1\times 10^{8}\ Pa[/tex]

The pressure at which the finite volume of oxygen molecules causes deviations from ideal-gas behavior can be estimated using the ideal gas law and considering the volume of a molecule. At ordinary temperatures, the approximate pressure is 2.04 × 10^11 Pa.

To estimate the pressure at which the finite volume of oxygen molecules causes noticeable deviations from ideal-gas behavior, we need to equate the volume of gas per molecule to the volume of a molecule. The volume of the gas per molecule is equivalent to the volume of a sphere with a radius of 2.0 × 10-10 m, which is approximately 4.19 × 10-30 m3. Using the ideal gas law, we can calculate the pressure using the formula PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Assuming 1 mole of oxygen gas (which contains 6.022 × 1023 molecules), the number of moles can be calculated by dividing the volume of the gas per molecule by Avogadro's number. The approximate pressure at which deviations from ideal-gas behavior become noticeable is around 2.04 × 1011 Pa (Pascals).

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