Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows at a constant acceleration. While coming to a stop, the driver looks in the rear-view mirror and notices that the box is not slipping. In what direction is the frictional force acting on the box?

Answer:

The critical angle is 41.8°.

Explanation:

The critical angle is [tex]\theta_1[/tex] for which the angle of refraction [tex]\theta_2[/tex] is 90°. From Snell's law we have

[tex]n_1sin (\theta_1 ) = n_2 sin(\theta_2)[/tex]

[tex]n_1sin (\theta_1 ) = n_2 sin(90^o)[/tex]

[tex]sin (\theta_1 ) = n_2/n_1,[/tex]

[tex]\theta_1 = sin^{-1}(\dfrac{n_2}{n_1} ).[/tex]

Putting in [tex]n_2 =1.00,[/tex] and [tex]n_1 = 1.50[/tex] we get:

[tex]\theta_1 = sin^{-1}(\dfrac{1.00}{1.500} ),[/tex]

[tex]\boxed{\theta_1 = 41.8^o}[/tex]

Thus, the critical angle is 41.8°.

Final answer:

The critical angle for light propagating from a material with an index of refraction of 1.50 to a material with an index of refraction of 1.00 is approximately 41.8 degrees.

Explanation:

The critical angle θcrit for light propagating from a material with an index of refraction (n1) of 1.50 to a material with an index of refraction (n2) of 1.00 is found using Snell's Law and the concept of total internal reflection. The formula to calculate the critical angle is θcrit = sin−1(n2/n1). Plugging in the values for the indices of refraction, we have θcrit = sin⁻¹(1.00/1.50).

Performing the calculation, we get:

θcrit = sin⁻¹(0.6667) ≈ 41.8°

Therefore, the critical angle for light traveling from a medium with an index of refraction of 1.50 to a medium with an index of refraction of 1.00 is approximately 41.8 degrees.

Answer:

[tex]\eta_{th} = 8.247\%[/tex]

Explanation:

The maximum possible efficiency for the floating power plant is given by the Carnot's Efficiency:

[tex]\eta_{th} = \left(1-\frac{278.15\,K}{303.15\,K} \right)\times 100\%[/tex]

[tex]\eta_{th} = 8.247\%[/tex]

Answer:

[tex]W_f=-62460\ J[/tex]

Explanation:

Given that

mass of the car ,m = 120 kg

Radius ,R= 12 m

Speed at the bottom , u = 25 m/s

Speed at top ,v= 8 m/s

We know that

Work done by all the forces = Change in the kinetic energy

Work done by gravity +  Work done by friction =Change in the kinetic energy

By taking point A as reference

[tex]m g \times (2R) + W_f=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2[/tex]

Now by putting the values in the above equation we get

[tex]120\times 10\times 2\times 12+ W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2[/tex]

[tex]W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2-120\times 10\times 2\times 12\ J[/tex]

[tex]W_f=-62460\ J[/tex]

Therefore the work done by friction force will be -62460 J.

Answer:

58.8 N

Explanation:

Let 'F₁' be the force by front arm and 'F₂' be the force by back arm.

Given:

Mass of the rod (m) = 3.00 kg

Length of the pole (L) = 4.50 m

Acceleration due to gravity (g) = 9.8 m/s²

Distance of 'F₁' from one end of pole (d₁) = 0.750 m

'F₂' acts on the end. So, distance between 'F₁' and 'F₂' = 0.750 m

Now, weight of the pole acts at the center of pole.

Now, distance of center of pole from 'F₁' is given as:

d₂ = (L ÷ 2) - d₁

[tex]d_2=\frac{4.50}{2}-0.75=1.5\ m[/tex]

Now, as the pole is held horizontally straight, the moment about the point of application of force 'F₁' is zero for equilibrium of the pole.

So, Anticlockwise moment  = clockwise moment

[tex]F_2\times d_1=mg\times d_2\\\\F_2=\frac{mg\times d_2}{d_1}[/tex]

Plug in the given values and solve for 'F₂'. This gives,

[tex]F_2=\frac{3.00\ kg\times 9.8\ m/s^2\times 1.5\ m}{0.75\ m}\\\\F_2=\frac{44.1}{0.75}=58.8\ N[/tex]

Therefore, the force exerted by the back arm on the pole is 58.8 N vertically down.

Answer:

[tex]L=4.8*10^{17}m[/tex]

Explanation:

Given data

Power P=4.00×10²W

Radius r=6.5×10⁴m

Voltage V=120V

To find

Length of wire L

Solution

We know that resistance of wire can be obtained from

[tex]P=\frac{V_{2}}{R}\\ R=\frac{V_{2}}{P}[/tex]

We also know that R=pL/A solving the length noting that A=πr²

and using p=100×10⁻⁸Ω.m we find that

So

[tex]L=\frac{RA}{p}\\ L=\frac{\frac{(V^{2})}{P}(\pi r^{2}) }{p} \\L=\frac{V^{2}(\pi r^{2})}{pP}\\ L=\frac{(120V)^{2}\pi (6.5*10^{4} m)^{2} }{100*10^{-8}(4.00*10^{2} W) }\\ L=4.8*10^{17}m[/tex]

Final answer:

To calculate the length of Nichrome wire needed for a heating element using 400 W at 120 V, one determines the resistance and then uses it with the wire's cross-sectional area and resistivity. Approximately 41.45 meters of wire is required.

Explanation:

To estimate the necessary length of Nichrome wire for a laboratory heater, we start by calculating the resistance using the power and voltage supplied. The formula for power (P) in terms of voltage (V) and resistance (R) is P = V2 / R. Given P = 400 W and V = 120 V, the resistance can be calculated as follows:

R = V2 / P = (1202) / 400 = 36 Ω.

Next, we use the resistivity (ρ) of Nichrome and the cross-sectional area (A) of the wire to find the length (L). The formula R = ρL / A is applicable here, where ρ for Nichrome is approximately 1.10×10-6 Ω·m and A = πr2. The radius (r) given is 6.5×10-4 m, so:

A = π(6.5×10-4)2 = 1.33×10-6 m2.

Substituting the values into the formula, we get:

L = (R · A) / ρ = (36 · 1.33×10-6) / (1.10×10-6) ≠ 41.45 m.

Thus, approximately 41.45 meters of Nichrome wire is required for the heater to use 400 W of power at 120 V.

Answer:

note:

please find the attachment

Explanation:

A frayed cord means that the insulation of the cord is worn out which exposes the cord and when it touches any other conductor a short circuit happens which is basically the flow of very high amount of current.

If we talk about the example of frayed cord, when such short circuit happens the lamp will not turn on and a massive amount of fault current will flow throughout the path of short circuit.

Now what happens?

There are two possibilities;

Answer: In a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point.

Explanation: This law was put forward by Blaise Pascal, a French mathematician in 1648. Pascal's Law states that in a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point. Pascal's law has been used in fluid mechanics for different applications these includes:

- the hydraulic jack used in automobile listings,

- most automobile braking systems,

-water towers, and dams.

Answer:

Explanation:

Given an RC series circuit

Initially uncharged capacitor

C=3.67 μF

Resistor R=8.01 k Ω=8010 ohms

Battery EMF(V)=1.5V with negligible internal resistance.

The initial current in the circuit?

At the beginning the capacitor is uncharged and it has a 0V, so all the voltage appears at the resistor,

Now using ohms law

V=iR.

i=V/R

i=1.5/8010

i=0.000187A

1mA=10^-3A

Therefore, 1A = 1000mA

i=0.187 milliamps

The initial current in the circuit is 0.187 mA

Answer:

The initial current is 0.0187 mA.

Explanation:

Given that

capacitance is given as 3.67 x 10⁻⁶ F

resistance is given as 8010 Ω

voltage across the circuit is 1.5 V

Since the capacitor is initially uncharged, the capacitive reactance is zero.

From ohms law;

Voltage across the circuit is directly proportional to the opposition to the flow of current.

In these circumstances, as the battery only "sees"a resistor, the initial current can be found applying Ohm's law to the resistor, as follows:

[tex]V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{8.011e3\Omega}\\ = 0.0187 mA[/tex]

The initial current (that will be diminishing as the capacitor charges), is 0.0187 mA.

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

70.31kg

Step I: Consider Newton's second law of motion which states that;

∑F = m x a;

Where;

∑F = net force acting on a body

m = the mass of the body

a = acceleration due to the force on the body.

Step II: Now to the question;

Since frictional force is negligible and the only force acting on the block of ice is the applied force by the dockworker, the net force on the body (block of ice) is the constant horizontal force. i.e

∑F = 90.0N

Also;

the block starts from rest and moves a distance (s) of 13.0m in a time (t) of 4.50s. Here, we can get the acceleration in that duration of time using one

the equations of motion as follows;

s = ut + [tex]\frac{1}{2}[/tex]at²            ------------------------------(ii)

Where;

s = distance covered = 13.0m

u = initial velocity = 0      [since the block starts from rest]

t = time taken to cover the distance = 4.50s

a = acceleration of the body.

Substitute these values into equation (ii) as follows;

13.0 = 0(4.5) + [tex]\frac{1}{2}[/tex](a)(4.50)²

13.0 = 0 + [tex]\frac{1}{2}[/tex](a)(20.25)

13.0 = [tex]\frac{1}{2}[/tex](a)(20.25)

13.0 = 10.125a

Solve for a;

a = [tex]\frac{13.0}{10.125}[/tex]

a = 1.28m/s²

Step III: Now substitute the values of a = 1.28m/s² and ∑F = 90.0N into equation (i) as follows;

90.0 = m x 1.28

m = [tex]\frac{90.0}{1.28}[/tex]

m = 70.31

Therefore, the mass of the block of ice is 70.31kg

Answer:

Explanation:

A force is provided by the magnetic field which perpendicular to both the velocity of the charge and magnetic field

F = qvB  where B is the magnetic field in tesla, q is charge and v is velocity

The potential energy is transferred into kinetic energy

PE = Vq = 1/2 mv²

v = √(2Vq/ m)

charge on the proton = 1.602 × 10 ⁻¹⁹ C and mass of a proton = 1.673 × 10⁻²⁷

v = √ (( 2 × 1.602 × 10 ⁻¹⁹ C × 0.75 × 10³V ) / (1.673 × 10⁻²⁷)) = √( 1.4363 × 10¹¹ ) = 3.79 × 10⁵ m/s

The force of magnetic field produces centripetal force

qvB = mv² /R

where  R radius = 1.80mm = 0.0018 m / 2 = 0.0009 m

qvB = ( m / R ) × (2qV /m)

cancel the common terms

vB = 2V / R

3.79 × 10⁵ m/s × B = 2 × 0.75 × 10³V / 0.0009 = 1.667 × 10⁶

B = 1.667 × 10⁶ / 3.79 × 10⁵ m/s = 4.40 T

b) magnetic field needed for the electron

qvB = mv² /R where m is the mass of an electron = 9.11 × 10⁻³¹ Kg

qB = mv/R

qB = ( 9.11 × 10⁻³¹ Kg × 3.79 × 10⁵ m/s) / 0.0009

qB  = 3.8363 × 10 ⁻²²

B = 3.8363 × 10 ⁻²² / 1.602 × 10⁻¹⁹ kg = 0.0239 T

Answer:

Magnetic energy stored in the inductor when all of the energy in the circuit is in the inductor = 0.049 mJ

If all the energy is then transferred into the capacitor, the voltage drop across the capacitor = 0.00572 V = 0.01 V (expressed to the hundredths value)

Explanation:

In an RLC circuit with maximum current of 7mA = 0.007 A

When all of the energy is stored in the inductor, maximum current will flow through it,

Hence E = (1/2) LI²

L = inductance of the inductor = 2 H

E = (1/2) (2)(0.007²) = 0.000049 J = 0.049 mJ

When all the energy in the circuit is in the capacitor, this energy will be equal to the energy calculated above.

And for a capacitor, energy is given as

E = (1/2) CV²

E = 0.000049 J, C = 3 F, V = ?

0.000049 = (1/2)(3)(V²)

V = 0.00572 V = 0.01 V

The velocity at which the long jumper landed can be computed by separating the jump into horizontal and vertical components, using the initial take-off angle and the horizontal distance. The horizontal and vertical velocities are combined to find the magnitude of the total landing velocity, and the arctangent of their ratio gives the direction.

To determine the velocity at which the long jumper landed, we need to consider the projectile motion of the jump. There are two components of velocity to consider: the horizontal (vx) and the vertical (vy) at the point of landing.

First, the horizontal velocity (vx) can be found by dividing the total horizontal distance by the time of flight (t). The equation of horizontal motion is:

vx = d / t, where d is the horizontal distance.

The vertical velocity (vy) of the jumper when he lands will be the same magnitude but opposite in direction to the vertical velocity at take-off due to symmetry in projectile motion, assuming no air resistance. The vertical velocity at take-off can be calculated using the initial jump angle (Ө) and the initial velocity (v0).

Using the initial angle of 30°, the vertical component of the initial velocity at take-off (v0y) is:

v0y = v0 * sin(Ө).

Since the vertical motion is subject only to acceleration due to gravity (g), the final vertical velocity at landing (vy) will be v0y (but in the opposite direction).

The total landing velocity (v) is then found by combining these two components using the Pythagorean theorem:

v = √(vx^2 + vy^2).

The direction of the landing velocity is given by the angle made with the horizontal, found using the arctangent of the ratio of vy over vx:

θ = arctan(vy / vx).

These calculations would provide the magnitude and direction of the landing velocity.

Answer:

981.41 secs

Explanation:

Parameters given:

Voltage, V = 1.5V

Power, P = 0.204W

Number of electrons, n = 8.33 * 10^20

First, we calculate the current:

P = I*V

I = P/V

I = 0.204/1.5 = 0.136A

The total charge of 8.33 * 10^20 electrons is:

Q = 8.33 * 10^20 * 1.6023 * 10^(-19)

Q = 133.47 C

Current, I, is given as:

I = Q/t

=> t = Q/I

t = 133.47/0.136 = 981.41 secs

In physics, power is calculated using the formula P = IV, where P is power, I is current, and V is voltage. By applying this formula, along with the relationship between charge, current, and time, the time the flashlight was used can be calculated.

Power is determined by the rate at which energy is transferred, calculated as P = IV where P is power, I is current, and V is voltage. In this scenario, the power supplied is 0.204 W from a 1.50-V battery. The formula P = IV can be rearranged to find the current flowing, which is 0.136 A.

To determine the time the flashlight was used, we can utilize the relationship between charge, current, and time: Q = It. Given 8.33 x 10^20 electrons moved, we need to convert this to Coulombs by recognizing that 1 electron has a charge of approximately 1.6 x 10^-19 C. By dividing the total charge by the current, we find the time t to be approximately 1.23 x 10^19 seconds.

Answer

[tex]3.9*10^{5}N/C[/tex]

Explanation:

from the expression for determing the magnetic field strength

[tex]E=\frac{q}{4\pi e_{0}d^2 }\\[/tex]

since the charge is given as

[tex]q=1.56*10^{-5}c\\[/tex]

and the distance is

d=60cm=0.6m

We can calculate the constant k

[tex]K=\frac{1}{4\pi e_{0}}\\ K=\frac{1}{4\pi *8.8542*10^{-12}}\\k=8.98*10^{9}\\[/tex]

if we substitute values, we arrive at

[tex]E=\frac{8.98*10^9 *1.56*10^{-5}}{0.6^2} \\E=389133.33\\E=3.9*10^{5}N/C[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

Ф = 4.98 °

Explanation:

Given:

R = 1.75 km = 1.75 * 10^3

g = 9.8 m/s^2

v = 120 km/h ==> 33.33 m/s

required:

Ф

solution:

The angle(Ф) is given by

Ф = arctan(v^2/R*g)

   = arctan(33.33^2/9.8*1.75 * 10^3)

   = 4.98 °

note:

there maybe error in calculation but method is correct.

Answer:

The Reasons for which petroleum is the chosen fuel for transportation in the U.S. include number 1) and 2).

Explanation:

The main reasons why oil is the most used fuel in the USA is because of its energy potential with a high combustion power and a better octane rating compared to other fuels. Thus, fossil fuel derived from oil has a high energy value per unit volume and an ability to quickly start or stop providing energy.

Answer:

The answer to this question is attached.

The pressure gradient required to accelerate the air inside a pipe can be calculated using Euler's simplified equation for fluid motion. At standard temperature and pressure, you would need a pressure gradient of approximately -102.3 Pa/m to accelerate air at 274 ft/s².

To calculate the pressure gradient, or dp/ds, required to accelerate the air inside a pipe, we can use Euler's equation for fluid motion, which in its simplified form is dp/ds = -ρa, where ρ represents the density of the fluid (in this case, air) and a is the acceleration. For standard temperature and pressure, the density of the air is approximately 1.225 kg/m³. It is important to note that the acceleration provided is in feet per second squared, so we would need to convert that to meters per second squared to match units with the density. The conversion factor is 1 ft/s² = 0.3048 m/s², resulting in an acceleration of approximately 83.515 m/s². By substituting these values into the equation, we get dp/ds = -(1.225 kg/m³)(83.515 m/s²) = -102.3 kg/(m·s²), or -102.3 Pa/m, since a Pascal (Pa) is equivalent to a kg/(m·s²).

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Answer: 0.03 N/C

Explanation:

We use the current density formula to solve this question.

I/A = σ * E

Where,

I = current flowing in the circuit = 0.3 A

A = cross sectional area of the wire = 1 mm²

σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1

E = strength of the electric field required

I/A = σ *E

E = I/(A * σ)

First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m

E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)

E = 0.3 A / 10 Ω^-1

E = 0.03 N/C

The required electric field to drive given current through the wire is 0.03 V/m.

To calculate the strength of the electric field required to drive a current of 0.3 amperes through a platinum wire, we can use the relationship between current density, electrical conductivity, and electric field.

The current density (J) is given by:

J = I / A

where I = 0.3 A and A = 1 mm² = 1 × 10⁻⁶ m².

Substituting the values, we get:

J = 0.3 A / (1 × 10⁻⁶ m²) = 3 × 10⁵A/m²

Next, we use Ohm's law in the form that relates current density to the electric field:

J = σE

where σ is the electrical conductivity of platinum, σ = 1.0 × 10⁷ S/m (S = 1/Ω), and E is the electric field.

Rearranging for the electric field, we get:

E = J / σ

Substituting the values, we get:

E = (3 × 10⁵ A/m²) / (1.0 × 10⁷ S/m) = 3 × 10⁻² V/m

Therefore, the strength of the very small electric field required to drive the current through the wire is 0.03 V/m.

Answer:

intensity at a point on the circle at an angle of 4.45° from the center line is 1.77 W/m².

Explanation:

See attached picture.

The intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².

To find the intensity at a point on the circle at an angle of 4.45° from the centerline, we need to consider the interference pattern created by the two transmitters.

1. Wavelength Calculation:

The wavelength [tex](\(\lambda\))[/tex] of the transmitted signal can be calculated using the frequency [tex](\(f\))[/tex] and the speed of light c.

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

Given:

- [tex]\( c = 3 \times 10^8 \)[/tex] m/s (speed of light)

- [tex]\( f = 1575.42 \times 10^6 \)[/tex] Hz (frequency)

[tex]\[ \lambda = \frac{3 \times 10^8}{1575.42 \times 10^6} \approx 0.1902 \text{ meters} \][/tex]

2. Phase Difference Calculation:

The phase difference [tex](\(\Delta \phi\))[/tex] between the two waves at a point on the circle is determined by the path difference [tex](\(\Delta L\))[/tex] travelled by the waves.

The path difference is given by:

[tex]\[ \Delta L = d \sin(\theta) \][/tex]

where:

- d is the distance between the two transmitters (5.18 mm = 0.00518 m)

- [tex]\(\theta\)[/tex] is the angle from the centerline (4.45°)

[tex]\[ \Delta L = 0.00518 \sin(4.45^\circ) \approx 0.00518 \sin(0.0776) \approx 0.00518 \times 0.0775 \approx 0.0004019 \text{ meters} \][/tex]

The phase difference is:

[tex]\[ \Delta \phi = \frac{2 \pi \Delta L}{\lambda} = \frac{2 \pi \times 0.0004019}{0.1902} \approx 0.0133 \text{ radians} \][/tex]

3. Intensity Calculation:

The intensity at a given point due to two sources interfering constructively or destructively can be found using:

[tex]\[ I = I_0 \left(1 + \cos(\Delta \phi)\right) \][/tex]

Here, [tex]\(I_0\)[/tex] is the intensity when both waves interfere constructively at the centerline [tex](\(\theta = 0^\circ\))[/tex], which is given as 2.00 W/m^2.

[tex]\[ I = 2.00 \left(1 + \cos(0.0133)\right) \][/tex]

Since [tex]\(\cos(0.0133) \approx 0.9999\)[/tex]:

[tex]\(\cos(0.0133) \approx 0.9999\)[/tex]

Thus, the intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².

Answer:

The projectile is in air for 13.03 seconds.

Explanation:

Given that,

Angle of projection of the projectile, [tex]\theta=55^{\circ}[/tex]

Initial speed of the projectile, u = 78 m/s

To find,

We need to find the time of flight of the projectile.

Solution,

It is defined as the time taken by the projectile when it is in air. It is given by the formula as :

[tex]T=\dfrac{2u\ \sin\theta}{g}[/tex]

[tex]T=\dfrac{2\times 78\ \sin(55)}{9.8}[/tex]

T = 13.03 seconds

So, the projectile is in air for 13.03 seconds.

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ([tex]C_{HP}[/tex]) is given by

[tex]C_{HP} = \dfrac{Q_{H}}{W_{in}}[/tex]

where, [tex]Q_{H}[/tex] is the magnitude of heat transfer between cyclic device and    high-temperature medium at temperature [tex]T_{H}[/tex] and [tex]W_{in}[/tex] is the required input and is given by [tex]W_{in} = Q_{H} - Q_{L}[/tex], [tex]Q_{L}[/tex] being magnitude of heat transfer between cyclic device and low-temperature [tex]T_{L}[/tex]. Therefore, from above equation we can write,

[tex]&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}[/tex]

Given, [tex]T_{L} = 460 K[/tex] and [tex]T_{H} = 540 K[/tex]. So,  the minimum work per unit heat transfer is given by

[tex]\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15[/tex]

The minimum work required to operate a heat pump between two thermal reservoirs at 460 K and 540 K is determined using the performance coefficient equation, Kp = Qh/W = Th/(Th – Tc). The efficiency of a heat pump is determined by the energy it transfers through heat and the input work required. To improve efficiency, the temperature difference between the hot and cold reservoir should be maximized.

The question pertains to the work-power balance in a heat pump operating between two thermal reservoirs. It involves the concept of thermal efficiency and the performance coefficient (Kp) of the heat pump.

Given the heat pump's thermal energy reservoirs at 460 K and 540 K, one can find the minimum work using the equation Kp = Qh/W = Th/(Th – Tc). Here, Th and Tc are the temperatures of the hot and cold reservoirs, respectively, Qh is the heat delivered to the hot reservoir, and W is the work done.

Such a heat pump operates on the principle of heat transfer of energy from a low-temperature reservoir to a high-temperature one, which requires input work. The quality of a heat pump is judged by the energy transferred by heat into the hot reservoir and the input work required.

The conservation of energy is not violated in this process, as the heat pump might extract energy from the ambient air or ground, contingent on its settings. Also, to improve the efficiency of the heat pump, the temperature of the hot reservoir should be elevated, and the cold reservoir should be lowered as per the Carnot efficiency equation.

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Answer:

the final temperature of the system of the system is 25.32°C

Explanation:

We are not given specific capacity of water and aluminium, so we use their standard values, also we are not given the density if water so we assume the standard vale of density of water

The aluminium calorimeter has a mass Mc= 120g

Volume of water in calorimeter = 100ml at θc =20°C

Density of water is

1000Kg/m³ = 1g/mL

Then, density = mass/ volume

Mass=density ×volume

Mass=1g/mL×100mL

Mass=100gram

Then, the mass of water is

Mw = 100gram

Mass of brass is Mb = 100gram

The temperature of brass is θb=100°C

The specific heat capacity of water is Cw= 1cal/g°C

The specific heat capacity of aluminum Ca=0.22cal/g°C

We are looking for final temperature θf=?

Given that the specific heat capacity of brass is Cb=0.09Cal/g°C

Using the principle of calorimeter;

The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.

So, the calorimeter gained heat and the liquid in the calorimeter gain heat too

Heat gain by calorimeter(Hc) = Mc•Ca•∆θ

Where Mc is mass of calorimeter,

Ca is Specific Heat capacity of Calorimeter

∆θ=(θf-θc)

Hc=Mc•Ca•∆θ

Hc=120•0.22•(θf-20)

Hc=26.4(θf-20)

Hc=26.4θf-528

Also, heat gain by the water

Heat gain by wayer(Hw) = Mw•Cw•∆θ

Where Mw is mass of water,

Cw is Specific Heat capacity of water

∆θ=(θf-θw),

Note that the temperature of the water and the calorimeter are the same at the beginning i.e. θc=θw=20°C

Hw=Mw•Cw•∆θ

Hw=100•1•(θf-20)

Hw=100(θf-20)

Hw=100θf-2000

Also heat loss by the brass is given by

heat loss by brass

Heat loss by brass(Hb)= Mb•Cb•∆θ

Where Mb is mass of brass,

Cb is Specific Heat capacity of brass

∆θ=(θb-θf)

Therefore,

Hb=Mb•Cb•∆θ

Hb=100•0.09•(100-θf)

Hb=9(100-θf)

Hb=900-9θf

Applying the principle of calorimeter

Heat gain = Heat loss

Hc+Hw=Hb

26.4θf-528 + 100θf-2000=900-9θf

26.4θf+100θf+9θf=900+2000+528

135.4θf=3428

Then, θf=3428/133.4

θf=25.32°C

The final temperature of the system is 25.32 degree Celsius.

Given data:

The mass of aluminum cup is, m = 120 g .

The mass of brass piece is, m' = 100 g.

The volume of water in aluminum cup is, V = 100 ml.

The temperature of water is, T = 20 degree Celsius.

The specific heat of brass is, [tex]c''=0.09 \;\rm cal/g ^\circ C[/tex].

The temperature of brass is, T'' = 100 degree Celsius.

The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.

So, first calculate heat gain by calorimeter,

H = mc (T' - T)

Here, c is the specific heat of aluminum and its value is, [tex]0.22 \;\rm cal/g^\circ C[/tex]. Solving as,

[tex]H = 120 \times 0.22 (T' - 20)\\H = 26.4T' - 528[/tex]

And, heat gain by water is,

[tex]H'=m'c'(T'-T)[/tex]

Here, c' is the specific heat of water. ( c' =1 )

Solving as,

[tex]H'=100 \times 1 \times (T'-20)\\H' = 100T' -2000[/tex]

Now, heat loss by brass is,

[tex]H'' = m'c'' (T''-T')\\\\H'' = 100 \times 0.09 \times (100-T')\\\\H'' = 900-9T'[/tex]

Applying the principle of calorimeter

Heat gain = Heat loss

H + H' = H''

(26.4T' - 528) + (100T' - 2000) = (900 - 9T')

135.4 T' = 3428

T ' = 25.32 degree Celsius

Thus, we can conclude that the final temperature of the system is 25.32 degree Celsius.

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Answer:

(a) The initial speed required is 13116 m/s

(b) The escape speed is 10394 m/s

This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.

The full calculation can be found in the attachment below.

Explanation:

In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.

The calculation can be found below.

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁ [tex]ln[\frac{P_{1} }{P_{2} }][/tex]

   = mRT₁[tex]ln[\frac{P_{1} }{P_{2} }][/tex]      [∵pV = mRT]

   = (5) (0.287) (272.039) [tex]ln[\frac{2.0}{1.0}][/tex]

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU = [tex]mc_{v}(T_{2} - T_{1} ) = 0[/tex]

From 1st law of thermodynamics

Q = ΔU  + W

   = 0 + 270.588

   = 270.588 kJ

To find the work and heat transfer in this process, we can apply the First Law of Thermodynamics. We can determine the work done using the equation pV = constant and calculate the heat transfer using the equation Q = CpdT.

To find the work and heat transfer in this process, we need to apply the First Law of Thermodynamics. The First Law states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system. In this case, since the process is isobaric (constant pressure), we can determine the work done by integrating the equation pV = constant. This will give us the equation W = p(Vf - Vi), where W is the work done, p is the pressure, and Vf and Vi are the final and initial volumes respectively.

Since the process is isobaric, the heat transfer can be calculated using the equation Q = CpdT, where Q is the heat transfer, Cp is the specific heat at constant pressure, and dT is the change in temperature. In this case, since the temperature change is not given explicitly, we can assume it to be the same as the change in internal energy, which gives us dT = dU/Cp.

By substituting the given values into the equations, we can calculate the work and heat transfer.

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Answer:

TRUE. The potential of a negatively charged conductor must be negative

Explanation:

Let's examine each statement

The positively charged proton moves in the direction of the electric field, the power and the electric field are related

                ΔU = - E ds

                [tex]U_{f}[/tex] - U₀ = - E ds

                E = (U₀ –U_{f}) / s

To have a positive electric field the initial potential must be greater than the final potential, so the proton moves from a greater potential to a smaller one.

This statement is FALSE

The second statement

The potential has the same sign as the elective charge.

This statement is TRUE

A proton tends to move to a region of higher potential is False, while The potential of a negatively charged conductor must be negative is True.

The proton is a positively charged particle and moves in the direction of the electric field lines.

In the case of a positive charge, the electric field lines are away from the charge, which will push the proton away.

We know that the potential is inversely proportional to distance.

Thus, as the proton moves away, it is going from a higher potential to a lower potential.

The same can be proven in the case of an electric field generated by a negative charge, by using proper sign convention.

So the statement that a proton tends to go from a region of low potential to a region of high potential is FALSE

The potential has the same sign as the electric charge.

So the statement that the potential of a negatively charged conductor must be negative is TRUE

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Answer:

The number of polarizer needed so transmitted light has at least 12% intensity = 17

Explanation:

Given :

Angle between incident light and optic axis of polarizer = 20°

Given that, the transmission axis of each additional analyzer is rotated 20° relative to the transmission axis of the previous one

According to the malus law,

The intensity of the transmitted light passes through the polarizer is proportional to the square of the cosine of angle between the transmission axis to the optic axis.

⇒  [tex]I = I_{o} cos^{2} \alpha[/tex]

Where, [tex]I =[/tex] transmitted intensity through polarizer, [tex]I_{o} =[/tex] incident intensity of the light.

Given in question, all the time [tex]\alpha =[/tex] 20°

By calculation ∴ [tex]cos^{2} 20 = 0.883[/tex]

After 1st polarizer,

∴ [tex]I_{1} = 0.883I_{o}[/tex]

Now we need to multiply all the time 0.883 until we get 0.12 (relative 20° angle given in question)

After 17th polarizer we get 0.1205 ≅ 0.12

[tex]I_{17} = 0.883^{17} = 0.1205 \times 100 = 12[/tex]% [tex]I_{o}[/tex]

Means we get 12% intensity after 17th polarizing disk.

Answer:

L = 8.66 m

Explanation:

The length measured by the moving observer is related to the true length is given by

L = L₀ √1 - (v²/c²)

Where L₀ is the length of the spaceship as measured by person A, v is the speed of spaceship of person A and c is the speed of light c = 3.8x10ᵃ m/s

L = 10√1 - (0.5c)²/c²

L = 10√1 - (0.5*3.8x10ᵃ)²/3.8x10ᵃ ²

L = 8.66 m

Therefore, the Person B would measure the spaceship length to be 8.66 m

Answer:

Explanation:

Let the position of balance point be x distance from 1.5 µC on the right side . Balance point can not be on the left side or between the charges because in that case electric field by both the charges will be in the same direction .

For equilibrium of field

k x 4.75 µC / ( .35 +x )² -  k x  1.5  µC / x ² = 0

4.75 / ( .35 +x )² = 1.5 / x²

( .35 +x )² / x² = 4.75 / 1.5

( .35 +x )² / x² = 3.167

( .35 +x ) / x =  1.78

.35 + x = 1.78 x

.78 x = .35

x = .45 m ( + ve ) to the right of 1.50  µC

b ) When both the charges are positive , balance point will lie in between them because , electric field will be in opposite direction .

For equilibrium of field

k x 4.75 µC / ( .35 - x )² =   k x  1.5  µC / x ²

4.75 / ( .35 -x )² = 1.5 / x²

( .35 -x )² / x² = 4.75 / 1.5

( .35 -x )² / x² = 3.167

( .35 -x ) / x =  1.78

.35 -x = 1.78 x

2.78 x = .35

x = .126 m  ( - ve ) to the right of 1.50  µC

Answer:

Valid

Explanation:

to determine if the claim is valid, we compare the efficiency of the device to that of a Carnot engine.

The following data were given

High Temperature = 920R,

Low temperature =490R

work=4.5hp =4.5*2544.5=11450.25Btu/h

low heat Ql heat= 15000Btu/h

High heat Qh=work +Ql=11450.25Btu/h+15000Btu/h=26450.25Btu/h

Next we cal calculate the efficiency of the Carnot engine

[tex]E_Carnot=1-\frac{T_L}{T_H}\\ E_Carnot=1-\frac{490}{920}\\ E_Carnot=0.467[/tex]

Hence the maximum efficiency at the given temperature is 47%

Next we calculate the efficiency of the device

[tex]E_device=\frac{work}{Q_H} \\E_device=\frac{11450.25}{26450.25} \\E_device=0.433\\[/tex]

which is 43%

since the maximum efficiency of 47% is not exceeded, we can conclude that the claim is valid

Final answer:

The velocity and acceleration vectors of a baseball can change during its flight. At the ball's maximum height, its velocity is parallel to its acceleration. There is no point where the ball's velocity is perpendicular to its acceleration.

Explanation:

In a game of baseball, the velocity and acceleration of a ball can change as it moves through the air. (a) Yes, there is a point during the flight of the ball where its velocity is parallel to its acceleration. This occurs when the ball reaches its maximum height. At this point, the ball stops moving upward and starts moving downward. Both the velocity and acceleration vectors are directed downward and therefore parallel to each other.

(b) No, there is no point during the flight of the ball where its velocity is perpendicular to its acceleration. The orientation of the velocity and acceleration vectors will always be either parallel or antiparallel to each other.