If there were a great migration of people toward the Earth's equator, the length of the day would

a. decrease because of conservation of angular momentum.
b. remain unaffected.
c. increase because of conservation of angular momentum.
d. decrease because of conservation of energy.
e. increase because of conservation of energy.

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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The final temperature of the mixture is approximately 29.91°C.

The heat exchange between the tea and the ice. The temperature of the final mixture will be somewhere between 0°C and 31°C, and we need to determine that final temperature. The heat transfer can be calculated using the principle of conservation of energy:

Qin = Qout

The heat gained by the ice as it melts is given by:
Qice = m2 ⋅ Lf

The heat gained by the tea as it cools down is given by:
Qtea = m1 ⋅ c ⋅ (T1 - Tfinal)

The negative sign is used because the tea is losing heat. Setting these equal to each other and solving for Tfinal, we get:
m1 ⋅ c ⋅ (T1 - Tfinal) = m2 ⋅ Lf

Now, let's plug in the given values:

0.73 kg ⋅ 4186 J/(kg ⋅ K) ⋅ (31°C - Tfinal) = 0.095 kg ⋅ 33.5 × 10^4 J/kg

Now, solve for Tfinal:

0.73 ⋅ 4186 ⋅ (31 - Tfinal) = 0.095 ⋅ (33.5 × 10^4)

3050.78 ⋅ (31 - Tfinal) = 3182.5

94602.78 - 3050.78 ⋅ Tfinal = 3182.5

-3050.78 ⋅ Tfinal = -91420.28

Tfinal = 91420.28/3050.78

Tfinal ≈ 29.91°C

So, the final temperature of the mixture is approximately 29.91°C.

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

Final answer:

To determine the exit area of the nozzle, use the principle of conservation of mass and the equation for mass flow rate. Calculate the density using the Ideal Gas Law and substitute it into the equation for area.

Explanation:

To determine the exit area of the nozzle, we can use the principle of conservation of mass and the equation for mass flow rate:

Mass flow rate = density x velocity x area

Given that the mass flow rate is 2.5 kg/s and the velocity is 280 m/s, we can rearrange the equation to solve for the area:

Area = mass flow rate / (density x velocity)

However, we need to find the density of the steam at the nozzle exit. To do this, we can use the Ideal Gas Law:

Pressure x Volume = n x R x Temperature

Where pressure = 2 MPa, volume can be assumed to be the volume of the nozzle exit, R is the gas constant, and temperature is 400°C converted to Kelvin.

Once we have the density, we can substitute it into the equation for the area to find the exit area of the nozzle.

The exit area of the nozzle is approximately [tex]\( 0.00140 \text{ m}^2 \) or \( 1.40 \text{ mm}^2 \)[/tex].

The continuity equation for a steady-state flow is given by:

[tex]\[ \dot{m} = \rho \cdot A \cdot v \][/tex]

To find the density [tex]\( \rho \)[/tex], we need to use the ideal gas law, which is a good approximation for steam under these conditions:

[tex]\[ P = \rho \cdot R \cdot T \][/tex]

where:

- P is the absolute pressure at the nozzle exit (2 MPa or 2000 kPa),

- R is the specific gas constant for steam (0.4615 kJ/kg·K),

- T is the absolute temperature at the nozzle exit (400°C + 273.15 = 673.15 K).

Rearranging the ideal gas law to solve for [tex]\( \rho \)[/tex]:

[tex]\[ \rho = \frac{P}{R \cdot T} \][/tex]

Now, we can substitute the density [tex]\( \rho \)[/tex] back into the continuity equation to solve for the exit area A:

[tex]\[ A = \frac{\dot{m}}{\rho \cdot v} \][/tex]

Substituting the values we have:

[tex]\[ \rho = \frac{2000 \text{ kPa}}{0.4615 \text{ kJ/kg·K} \cdot 673.15 \text{ K}} \] \[ \rho = \frac{2000}{310.56} \text{ kg/m}^3 \] \[ \rho \approx 6.44 \text{ kg/m}^3 \][/tex]

Now, we can find the exit area A:

[tex]\[ A = \frac{2.5 \text{ kg/s}}{6.44 \text{ kg/m}^3 \cdot 280 \text{ m/s}} \] \[ A = \frac{2.5}{1787.2} \text{ m}^2 \] \[ A \approx 0.00140 \text{ m}^2 \][/tex]

Answer:

Current, 7.29 A

Explanation:

It is given that,

Length of the horizontal rod, L = 0.3 m

Magnetic field through a horizontal rod, [tex]B=6.4\times 10^{-2}\ T[/tex]

The magnetic force acting on the rod, F = 0.14 N

Let the current flowing through the rod is given by I. The magnetic force acting on an object in the uniform magnetic field is given by :

[tex]F=ILB\ sin\theta[/tex]

Here, [tex]\theta=90^{\circ}[/tex]

[tex]F=ILB[/tex]

[tex]I=\dfrac{F}{LB}[/tex]

[tex]I=\dfrac{0.14\ N}{0.3\ m\times 6.4\times 10^{-2}\ T}[/tex]

I = 7.29 A

So, the current flowing through the rod 7.29 A.

Final answer:

The current flowing in the rod is approximately 7.292 A when it is subject to a 0.140 N magnetic force within a 6.40 x [tex]10^-^2[/tex] T magnetic field. The calculation uses the formula F = I * L * B, and since the rod is perpendicular to the magnetic field, the angle θ is 90°, which simplifies the calculation.

Explanation:

The question asks for the current flowing through a rod that is experiencing a magnetic force due to an external magnetic field.

The force on a current-carrying conductor in a magnetic field is given by the equation F = I * L * B * sin(θ), where F is the force in newtons, I is the current in amperes, L is the length of the conductor in meters, B is the magnetic field in teslas, and θ is the angle between the direction of the current and the magnetic field. In this case, the rod is perpendicular to the magnetic field, so the angle θ is 90°, making sin(θ) equal to 1.

To find the current I, we rearrange the formula to be I = F / (L * B). Substituting the given values:

Force F = 0.140 N

Length L = 0.300 m

Magnetic field B = 6.40×[tex]10^-^2[/tex] T

The current I can thus be calculated as I = 0.140 N / (0.300 m * 6.40×[tex]10^-^2[/tex] T).

Performing the calculation, I equals approximately 7.292 A.

Answer:

D.The potential energy per unit charge

Explanation:

Electric potential of a charged particle:

It is scalar quantity because it has magnitude but it does not have direction.

It is the amount of work done required to move a unit positive charge from reference point to specific point in the electric field without producing any acceleration.

Mathematical representation:

[tex]V=\frac{W}{Q_0}[/tex]

Where W= Work done

[tex]Q_0[/tex]= Unit positive charge

Other formula to calculate electric field:

[tex]V=\frac{KQ}{r}[/tex]

Where K=[tex]\frac{1}{4\pi \epsilon_0}[/tex]

It can be defined as potential energy per unit charge.

Hence, option D is true.

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:

Explanation:

The energy of the particle in a square power well is given by the Schrödinger equation, with

       E₁ = RA (h²/ 8m L²)   n²

With h the Planck constant the mass of the particle, L the length of the box and n an integer starting 1

We find the energy for each situation presented

1) For the fundamental state n = 1

       E₁ = RA (h/ 8m L²)

2) the first excited state corresponds to n = 2

       E₂ = RA (h/ 8m L²) 2²

       E₂ = E₁ 4

3)    E₁ = RA (h/ 8m L²)

4) The length of the box is L = 2L

      E’= RA (h’ / 8m (2L)²)

      E’= RA (h’ / 8m L²) Ra ¼

      E’= E₁ / 2

5) L = 2l first excited state

      n = 2

      E ’’ = RA (h/ 8m (2L)²2) 2²

      E ’’ = RA (h ’/ 8m (L)²) 4/2

      E ’’ = E₁ 2

All energies are in relation to the fundamental state (E1), milking from least to greatest

       E1 / 2 <E1 = E1 <2E1 <4E1

       4 <3 = 1 <5 <2

Final answer:

The correct order from lowest to highest energy for a particle in a square well, considering both quantum state and box length, is: (5) < (4) < (3) < (2) < (1). This order is based on the quantum energy level formula E_n = n^2 * pi^2 * h-bar^2 / (2 * m * L^2), where energy increases with the square of the principal quantum number and decreases with the square of the box length.

Explanation:

To determine the proper order from lowest energy to highest energy for the particle in a square well, we must consider the principles of quantum mechanics, particularly the energy levels of a particle in a one-dimensional box. The energy levels are quantized and are given by the expression En = n2π2ħ2/(2mL2), where ħ is the reduced Planck constant, m is the mass of the particle, L is the length of the box, and n is the principal quantum number (n = 1 for the ground state, n = 2 for the first excited state, and so on). Given this, here's the reasoning for the order:

Therefore, the order from lowest energy to highest energy is as follows:

(5) < (4) < (3) < (2) < (1)

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:

The break even cost is $0.0063825

Explanation:

Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.

Wattage = W

Cost per kilo watt hour = C

Number of hours per year = H

Price per bulb/CFL = P

Discount rate = 11%

Life of bulb = 2 years

Price of bulb = $0.39

Wattage consumption of bulb per hours = 60

Life of CFL = 24 years

Price of CFL = $3.10

Wattage consumption of CFL per hour = 15

Calculate the Equated Annual Cost (EAC) of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)

PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods

PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)

Calculate the EAC of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

EAC = {-0.39 – (51.37570 x C)}/ 1.712523, consider this equation 1

Calculate the EAC of CFL

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, consider this equation 2

Equate 1 and 2 to find the amount of C

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

C = $0.0063825

Thus, the break- even cost per kilo – watt is $0.0063825

Answer: The silver pellet will release heat

Explanation:

Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.

Answer:

20.13841 rad/s²

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = [tex]500\times \frac{2\pi}{60}\ rad/s[/tex]

[tex]\omega_f[/tex] = Final angular velocity = 0

t = Time taken = 2.6 s

[tex]\alpha[/tex] = Angular acceleration

Equation of rotational motion

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-500\times \frac{2\pi}{60}}{2.6}\\\Rightarrow \alpha=-20.13841\ rad/s^2[/tex]

The magnitude of the angular acceleration of the CD, as it spins to a stop is 20.13841 rad/s²

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:

9.15 km/s

Explanation:

rp = 4.5 x 10^8 km

vp = 18.5 km/s

ra = 9.10 x 10^8 km

va = ?

According to the conservation of angular momentum constant.

Let m be the mass of planet

m x rp x vp = m x ra x va

4.5 x 10^8 x 18.5 = 9.10 x 10^8 x va

va = 9.15 km/s

To calculate the speed of the planet at apastron, we can use Kepler's second law and the given values of rp, vp, and ra. Plugging in the values, we find that the planet is moving at a speed of 0.92 km/s at apastron.

To calculate the speed of the planet at apastron, we can use Kepler's second law, which states that the area swept out by a planet in equal time intervals is constant. At periastron, the planet is moving fastest, so we can use the equation:

A1 = A2

where A1 is the area swept out at periastron and A2 is the area swept out at apastron.

Since the areas are equal, we can set up the following equation:

0.5 * rp * vp = 0.5 * ra * va

where rp is the distance at the periastron, vp is the velocity at the periastron, ra is the distance at the apastron, and VA is the velocity at the apastron. We can rearrange this equation to solve for va:

va = (rp * vp) / ra

Plugging in the given values, we get:

va = (4.50 x 10^8 km * 18.5 km/s) / (9.10 x 10^8 km)

va = 0.92 km/s

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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:

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

Explanation:

Answer:

[tex]\Delta P=1060184.8946\ Pa[/tex]

[tex]P_1=124651.2383\ Pa[/tex]

Explanation:

Given:

We know the Bernoulli's equation of in-compressible flow:

[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)

Cross sectional area of pipe at location 2:

[tex]A_2=\pi \frac{d_2^2}{4}[/tex]

[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]

[tex]A_2=0.0246\ m^2[/tex]

Cross sectional area of pipe at location 1:

[tex]A_1=\pi \frac{d_1^2}{4}[/tex]

[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]

[tex]A_1=0.0115\ m^2[/tex]

Using continuity equation:

[tex]A_1.v_1=A_2.v_2[/tex]

[tex]0.0115\times 9.83=0.0246\times v_2[/tex]

[tex]v_2=4.5953\ m.s^{-1}[/tex]

Now apply continuity eq. on both the locations:

[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]

[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]

[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]

[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)

Now the mass flow rate at location 1:

[tex]\dot{m_1}=\rho\times \dot{V}[/tex]

[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]

[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]

Now pressure at location 1:

[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]

[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]

[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)

The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as

dP = 114 kPa

Question Parameter(s):

Generally, the Bernoulli's equation   is mathematically given as

P + ρ*g*y + v² =pipe  constant

Where

A1*v1 = A2*v2

π*(0.105/2)²*9.91 = π*(0.167/2)²*v2

v2 = 3.9 m/s

Therefore

P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²

dP = 1290*9.8*9.01 + 9.91² - 3.9²

dP = 114 kPa

In conclusion, difference between fluid pressure is

dP = 114 kPa

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Answer:

mass percent of chlorine is 56.5%

Explanation:

Given data:

Volume V = 581 mL = 0.581 L

Pressure P = 752 mm of Hg = 752/760 = 0.9894 atm

Temperature T = 298 K

Molar gas constant R = 0.08206 atm.L/mol.K

from Ideal gas equation we have following relation

PV = nRT solving for n

[tex]n = \frac{PV}{RT}[/tex]

[tex]n = \frac{0.9894 \times 0.581}{0.08206 \times 298}[/tex]

n = 0.0235 mole

mas of [tex]Cl_2[/tex]  = moles × molar mass of [tex]Cl_2[/tex]

                        [tex]= 0.0235 \times 70.906 = 1.66 g[/tex]

Mass percent of chlorine  [tex]=  \frac{mass\ of\ Cl}{mass\ of\ sample}[/tex]

[tex]=\frac{1.66}{2.95} = 0.565 = 56.5 %[/tex]

Answer:

-0.31563 m/s

1.16437 m/s

Explanation:

[tex]m_1[/tex] = Mass of girl = 45 kg

[tex]m_2[/tex] = Mass of plank = 166 kg

[tex]v_1[/tex] = Velocity of girl relative to plank = 1.48 m/s

[tex]v_2[/tex] = Velocity of the plank relative to ice surface

In this system the linear momentum is conserved

[tex](m_1+m_2)v_2+m_1v_1=0\\\Rightarrow v_2=-\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=-\frac{45\times 1.48}{45+166}\\\Rightarrow v_2=-0.31563\ m/s[/tex]

Velocity of the plank relative to ice surface is -0.31563 m/s

Velocity of the girl relative to the ice surface is

[tex]v_1+v_2=1.48-0.31563=1.16437\ m/s[/tex]

The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

[tex]PE=\frac{GMm}{r}[/tex]

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

[tex]m = \rho V[/tex]

[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]

[tex]m = 7.210*10^{11}Kg[/tex]

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]

Then the potential energy at this point would be,

[tex]PE_1 = \frac{GMm}{r_1}[/tex]

[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]

[tex]PE_1 = 9.05*10^{15}J[/tex]

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]

The we calculate the Potential gravitational energy,

[tex]PE_2 = \frac{GMm}{r_2}[/tex]

[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]

[tex]PE_2 = 9.361*10^{15}J[/tex]

A biconvex lens with the given parameters is a converging lens. Using the Lens Maker's Equation with the radii of curvature and index of refraction for glass and air, the focal length of the lens is calculated to be approximately 12 cm.

A biconvex lens, where both surfaces of the lens bulge outwards, will bend light rays such that they converge at a focal point. With the parameters given (|R1|=10cm, |R2|=15cm, and nglass=1.5), we can deduce that this lens is a converging lens.

Part A: Since a biconvex lens makes parallel rays of light converge at a point after passing through the lens, it is classified as a converging lens.

Part B: To calculate the focal length (f) of the lens, we use the Lens Maker's Equation:

Carrying out the calculation with the data given (nglass=1.5, nair=1, R1=+10cm, and R2=-15cm), we get:

(1/f) = (1.5 - 1) ((1/10cm) - (1/(-15cm)))

(1/f) = 0.5 * (0.1cm⁻¹ + 0.0667cm⁻¹)

(1/f) = 0.5 * 0.1667cm⁻¹

(1/f) = 0.08335cm⁻¹

Therefore, the focal length f is the reciprocal of 0.08335cm⁻¹ which is approximately:

f ≈ 12cm

Answer:

m' = 1 x 10⁻⁶ kg/s

Explanation:

Given that

Diffussion constant = 1 x 10⁻¹¹

Mass flow rate ,m = 2 x 10⁻⁶ kg/s

The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.

So new flow rate m'

[tex]m'=\dfrac{m}{2}[/tex]

[tex]m'=\dfrac{2\times 10^{-6}}{2}\ kg/s[/tex]

m' = 1 x 10⁻⁶ kg/s

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:

-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

Answer:

a. [tex]F_2=31.76N[/tex]

b. [tex]F_2=185.86N[/tex]

Explanation:

Given:

[tex]F_1=27800N[/tex]

[tex]r_1=5.07x10^{-3}m[/tex]

[tex]r_2=0.150 m[/tex]

[tex]p=8.81x10^2 kg/m^3[/tex]

Using the equation to find the force so replacing

a.

[tex]F_1*A_2=F_2*A_1[/tex]

[tex]A=\pi*r^2[/tex]

[tex]F_2=F_1*\frac{A_2}{A_1}=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}[/tex]

[tex]F_2=31.76N[/tex]

b.

[tex]F_2=F_1+F_p[/tex]

[tex]F_2=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}+(8.81x10^2kg/m^3*9.8m/s^2*1.20m*\pi*(5.07x10^{-3})m^2)[/tex]

[tex]F_2=185.86N[/tex]

Answer:

1.6 penny

Explanation:

[tex]V_0[/tex] = Original volume = 1 gal (Assumed)

[tex]\Delta T[/tex] = Change in temperature

[tex]\beta[/tex] = Coefficient of volume expansion = [tex]280\times 10^{-6}\ /^{\circ}[/tex]

Change in volume is given by

[tex]\Delta_V=\beta V_0\Delta T\\\Rightarrow \Delta_V=280\times 10^{-6}\times 1\times (32-3.3)\\\Rightarrow \Delta_V=0.008036[/tex]

New volume would be

[tex]1+0.008036=1.008036\ gal[/tex]

The amount of money earned extra would be

[tex]0.008036\times 2=0.016072\ \$[/tex]

1.6 penny more would be earned if the temperature is 32°C

Final answer:

By refilling a container of apple cider at 32 degrees C instead of 3.3 degrees C, you would make approximately 1.6 pennies more per gallon due to thermal expansion of the cider.

Explanation:

To calculate how much more money you would make per gallon by refilling the container of apple cider when the temperature is 32 degrees C, as opposed to 3.3 degrees C, you need to determine the change in volume due to thermal expansion.

The formula for volume expansion is ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.

The initial temperature T1 is 3.3°C, and the final temperature T2 is 32°C, thus ΔT = T2 - T1 = 32°C - 3.3°C = 28.7°C. The coefficient of volume expansion of the cider, given as β, is 280 x 10^-6 (C°)^-1.

Assuming that the initial volume V₀ of the cider is 1 gallon, the change in volume ΔV would be:

ΔV = 280 x 10^-6 x 1 x 28.7 = 0.008036 gallons

To convert gallons to liters, we use the fact that 1 gallon is approximately 3.78541 liters. So, the increase in volume in liters would be:

ΔV (liters) = 0.008036 x 3.78541 = 0.0304 liters

Since there are approximately 3.78541 liters in a gallon, and knowing that the price for one gallon is two dollars, we can calculate the additional revenue (in pennies) as follows:

Extra revenue = ΔV (liters) / 3.78541 x 200 pennies = 0.0304 / 3.78541 x 200 ≈ 1.6 pennies

Therefore, you would make approximately 1.6 pennies more per gallon by refilling the container at 32°C compared to 3.3°C.

Answer:11.18 m/s

Explanation:

Given

mass of particle m=0.2 kg

Potential Energy U(x) is given by

[tex]U(x)=8x^2+2x^4[/tex]

at x=1 m

[tex]U(1)=8+2=10 J[/tex]

kinetic energy at x=1 m

[tex]K.E.=\frac{1}{2}mv^2=\frac{1}{2}\times 0.2\times 5^2[/tex]

[tex]K.E.=2.5 J[/tex]

Total Energy =U+K.E.

[tex]Total=10+2.5=12.5 J[/tex]

at x=0, U(0)=0

as total Energy is conserved therefore K.E. at x=0 is equal to Total Energy

[tex]\frac{1}{2}\times 0.2\times v^2=12.5[/tex]

[tex]v^2=125[/tex]

[tex]v=\sqrt{125}[/tex]

[tex]v=11.18 m/s[/tex]

By conserving energy, we find that the speed of the particle at the origin is 25 m/s.

The question is asking for the speed of the particle at the origin given its potential energy function and speed at x = 1.0 m. This can be solved using the principle of energy conservation which states that the total energy (kinetic energy + potential energy) of the particle is conserved unless acted upon by an outside force.

In this case, we first find the total energy of the particle at x = 1.0 m. The kinetic energy is (1/2)mv² = (1/2)*0.2 kg*(5 m/s)² = 2.5 J. The potential energy at x = 1.0 m, according to the given function, is U(1) = 8(1)² + 2(1)⁴ = 10 J. So, the total energy at x = 1.0 m is 2.5 J + 10 J = 12.5 J.

At the origin (x = 0), the potential energy is U(0) = 0. So, the kinetic energy at the origin is equal to the total energy, which is 12.5 J. From the kinetic energy, we can find the speed using the equation KE = (1/2)mv², which gives v = sqrt((2*KE)/m) = sqrt((2*12.5 J)/0.20 kg) = 25 m/s.

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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.

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.