When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner, which cools the entire house to 20°C in 38 min. If the COP of the air-conditioning system is 2.8, determine the power drawn by the air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air for which cv = 0.72 kJ/kg·°C and cp = 1.0 kJ/kg·°C

Answers

Answer 1
Final answer:

The power drawn by the air conditioner to cool the house from 35°C to 20°C in 38 minutes, given a COP of 2.8, is calculated to be 1.35 kW.

Explanation:

To calculate the power drawn by the air conditioner, we first determine the energy needed to cool the house from 35°C to 20°C. The energy required (Q) can be calculated using the formula Q = m⋅cv⋅ΔT, where m is the mass of the air, cv is the specific heat capacity at constant volume, and ΔT is the change in temperature.

Plugging the given values into the formula: Q = 800 kg ⋅ 0.72 kJ/kg°C ⋅ (35°C - 20°C) = 8,640 kJ.

Next, we use the Coefficient of Performance (COP) of the air conditioner to find the work input (W) required for this cooling process. The formula relating these quantities is COP = Q/W, rearranging for W gives W = Q/COP.

Therefore, W = 8,640 kJ / 2.8 = 3,085.71 kJ. To find the power drawn, we need the work in kilowatts (kW), knowing that 1 kW = 1 kJ/s, and that the total duration is 38 min or 2,280 seconds. Thus, the power drawn is Power = W / time = 3,085.71 kJ / 2,280 s = 1.35 kW.

Answer 2

Power drawn by the air conditioner is approximately 6.76 kW, calculated from COP, heat transfer, and time taken for cooling.

To solve this problem, we can use the energy balance equation for the air within the house:

[tex]\[ Q = mc\Delta T \][/tex]

where:

- [tex]\( Q \)[/tex] is the heat transfer,

- [tex]\( m \)[/tex] is the mass of air,

- [tex]\( c \)[/tex] is the specific heat capacity of air,

- [tex]\( \Delta T \)[/tex] is the change in temperature.

We also know that the Coefficient of Performance (COP) is defined as:

COP = Q_cooling/W_input

where:

- [tex]\( Q_{\text{cooling}} \)[/tex] is the heat removed from the house (in this case),

- W_input is the work input (power consumed) by the air conditioner.

We are given that the COP [tex](\( \text{COP} = 2.8 \))[/tex] and the initial and final temperatures [tex](\( T_{\text{initial}} = 35^\circ C \)[/tex], [tex]\( T_{\text{final}} = 20^\circ C \))[/tex].

First, let's calculate the heat transfer using the energy balance equation:

[tex]\[ Q = mc\Delta T \][/tex]

[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(35 - 20) \][/tex]

[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(15) \][/tex]

Q = 12000kJ

Now, we can find the work input using the COP formula:

COP = Q_cooling/{W_input

W_input = [tex]\frac{Q_{\text{cooling}}}{\text{COP}}[/tex]

W_input = [tex]\frac{12000 \, \text{kJ}}{2.8}[/tex]

W_input ≈ 4285.71 kJ

To find the power drawn by the air conditioner, we need to convert the energy to power. Since the time taken for cooling is 38 minutes, or [tex]\( \frac{38}{60} \)[/tex] hours:

Power = W_input /Time

[tex]\[ \text{Power} = \frac{4285.71 \, \text{kJ}}{\frac{38}{60} \, \text{hours}} \][/tex]

Power ≈ 6762.45 W

So, the power drawn by the air conditioner is approximately [tex]\( 6762.45 \, \text{W} \)[/tex] or [tex]\( 6.76 \, \text{kW} \)[/tex].


Related Questions

Write a balanced half-reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:  [tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Explanation:

Reduction is a process where electrons are gained and acidic solution means presence of [tex]H^+[/tex] ions.

Reduction of [tex]MnO_2[/tex] to [tex]Mn^{2+}[/tex]

Mn is in +4 oxidation state in [tex]MnO_2[/tex] which goes to +2 state in [tex]Mn^{2+}[/tex] by gain of 2 electrons.

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)[/tex]

In order to balance oxygen atoms:

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)+2H_2O[/tex]

In order to balance hydrogen atoms:

[tex]MnO_2(s)+4H^+(aq)\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

In order to balance charges:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l[/tex]

Thus the net balanced half reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Final answer:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, the balanced half-reaction is MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l).

Explanation:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, we can follow these steps:

1. Write the half-reaction for reduction, adjusting the physical states of the reactants and products:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

2. Balance the number of Mn and O atoms on each side of the equation:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

3. Balance the H atoms by adding H+ ions:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

4. Balance the charges by adding electrons:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

Therefore, the balanced half-reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

Part A: A charge +Q is located at the origin and a second charge, +9Q, is located at x= 15.8 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x -axis.

Part B: A charge +Q is located at the origin and a negative charge, -7Q, is located at a distance x= 19.6 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x-axis.

Answers

Answer:

Part a)

x = 3.95 cm

Part b)

x = - 11.9 cm

Explanation:

Part a)

Since both charges are of same sign

so the position at which net force is zero between two charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(15.8 - r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = 9Q[/tex]

[tex]\frac{Q}{r^2} = \frac{9Q}{(15.8 - r)^2}[/tex]

square root both sides

[tex](15.8 - r) = 3r[/tex]

[tex]r = 3.95 cm[/tex]

Part b)

Since both charges are of opposite sign

so the position at which net force is zero will lie on the other side of smaller charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(19.6 + r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = -7Q[/tex]

[tex]\frac{Q}{r^2} = \frac{7Q}{(19.6 + r)^2}[/tex]

square root both sides

[tex](19.6 + r) = 2.64r[/tex]

[tex]r = 11.9 cm[/tex]

so on x axis it will be at x = - 11.9 cm

What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m.s^{-1} ( i ) 9.5 × 10 5 m . s − 1 (ii)\:\:\:\:2.2\:\times\:10^9\:m.s^{-1} ( i i ) 2.2 × 10 9 m . s − 1 (iii)\:\:\:3.9\:\times\:10^8\:m.s^{-1} ( i i i ) 3.9 × 10 8 m . s − 1 (iv)\:\:\:\:1.7\:\times\:10^6\:m.s^{-1}

Answers

Answer:

Velocity of a proton, [tex]v=1.7\times 10^6\ m/s[/tex]    

Explanation:

It is given that,

Potential difference, [tex]V=15\ kV=15\times 10^3\ V[/tex]

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=qV[/tex]

q is the charge of proton

m is the mass of proton

[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}[/tex]

[tex]v=1695361.75\ m/s[/tex]

[tex]v=1.69\times 10^6\ m/s[/tex]

or

[tex]v=1.7\times 10^6\ m/s[/tex]

So, the velocity of a proton is [tex]1.7\times 10^6\ m/s[/tex]. Hence, this is the required solution.

Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29×1011 m.

Answers

Answer:

(a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

Explanation:

Given that,

Mass of baby = 4.20 kg

Mass of father = 100 kg

Distance = 0.200 m

We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times100}{0.200^2}[/tex]

[tex]F=7\times10^{-7}\ N[/tex]

(b). We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times1.9\times10^{27}}{(6.29\times10^{11})^2}[/tex]

[tex]F=1.35\times10^{-6}\ N[/tex]

Hence, (a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

The magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N

What is gravitational potential energy?

Gravitational potential energy is the energy which a body posses because of its position.

The gravitational potential energy of a body is given as,

[tex]G=\dfrac{Fr^2}{Mm}[/tex]

Here, (m) is the mass of the body, (F) is the gravitational force and (r) is the height of the body.

The mass of baby is 4.20 kg and mass of father is 100 kg. The distance between them is 0.200 m. Put the values in the above formula as,

[tex]6.67\times10^{-11}=\dfrac{F(0.200)^2}{100(4.20)}\\F=7\times10^{-7}\rm \; N[/tex]

(b) The magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29×1011 m.

Put the values in the above formula again as,

[tex]6.67\times10^{-11}=\dfrac{F(6.29\times10^{11})^2}{(1.9\times10^{27})(4.20)}\\F=1.35\times10^{-6}\rm \; N[/tex]

Thus, the magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N.

Learn more about the gravitational potential energy here;

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What energy is needed to raise the temperature of 2 kg of water from 20 ºC to 100 ºC. The heat capacity of water is 4190 J /kg/ ºC.

Answers

Answer:

670400 J

Explanation:

m = mass of water = 2 kg

T₀ = initial temperature of water = 20 ºC

T = initial temperature of water = 100 ºC

c = heat capacity of water = 4190 J /kg ºC

Q = energy needed to raise the temperature of water

Energy needed to raise the temperature of water is given as

Q = m c (T - T₀)

Inserting the values

Q = (2) (4190) (100 - 20)

Q = 670400 J

A 0.42 kg football is thrown with a velocity of 17 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.017 S. What is the force exerted on the receiver? Answer in units of N

Answers

Answer:

420 N

Explanation:

m = 0.42 Kg, u = 17 m/s, v = 0 m/s, t = 0.017 s

By first law of Newtons' laws of motion, the rate of change in momentum is force, F = m (v - u) / t

F = 0.42 x ( 0 - 17) / 0.017

F = - 420 N

Negative sign shows hat the force is resistive that means the ball finally comes to rest.

What is the best coefficient of performance for a refrigerator that cools an environment at -26.0°C and has heat transfer to another environment at 50.0°C?

Answers

Answer:

COP = 4.25

Explanation:

It is given that,

Cooling temperature, [tex]T_c=-26^{\circ}C=273-26=247\ K[/tex]

Heating temperature, [tex]T_h=50^{\circ}C=323\ K[/tex]

We need to find the coefficient of performance. It is given by :

[tex]COP=\dfrac{T_h}{T_h-T_c}[/tex]

[tex]COP=\dfrac{323}{323-247}[/tex]

COP = 4.25

So, the best coefficient of performance of a refrigerator is 4.45 Hence, this is the required solution.

What is the approximate energy required to raise the temperature of 1.00 L of hydrogen by 90 °C? The pressure is held constant and equal to 1 atm.

Answers

Answer:

Q = 116.8 J

Explanation:

Here given that the temperature of 1 L hydrogen is increased by 90 degree C at constant pressure condition.

So here we will have

[tex]Q = n C_p \Delta T[/tex]

here we know that

n = number of moles

[tex]n = \frac{1}{22.4}[/tex]

[tex]n = 0.0446[/tex]

for ideal diatomic gas molar specific heat capacity at constant pressure is given as

[tex]C_p = \frac{7}{2}R[/tex]

now we have

[tex]Q = (0.0446)(\frac{7}{2}R)(90)[/tex]

[tex]Q = 116.8 J[/tex]

A car starts from the origin and is driven 1.88 km south, then 9.05 km in a direction 47° north of east. Relative to the origin, what is the car's final location? a) Express your answer in terms of an angle (in degree)
b) and a distance.

Answers

Answer:

(a) θ = 55.85 degree

(b) 7.89 km

Explanation:

Using vector notations

A = 1.88 km south = 1.88 (- j) km = - 1.88 j km

B = 9.05 km 47 degree north of east

B = 9.05 ( Cos 47 i + Sin 47 j) km

B = (6.17 i + 6.62 j) km

Net displacement is

D = A + B

D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j

(a) Angle made with positive X axis

tanθ = 6.62 / 4.29 = 1.474

θ = 55.85 degree

(b) distance = [tex]Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}[/tex]

distance = 7.89 km

Final answer:

To find the car's final location, add the displacements in the north and east directions. The car's final location is 6.4 km north and 6.0 km east relative to the origin.

Explanation:

To find the car's final location, we need to add the displacements in both the north and east directions.

First, let's calculate the north displacement by using the distance formula:

North Displacement = 9.05 km * sin(47°) = 6.4 km

Next, let's calculate the east displacement by using the distance formula:

East Displacement = 9.05 km * cos(47°) = 6.0 km

Therefore, the car's final location relative to the origin is 6.4 km north and 6.0 km east.

A 10 kg mass starts from rest at the top of a frictionless incline and slides down the ramp. The ramp makes an angle of 0=30" . Take the positive x axis to be down the ramp. What is the acceleration of the mass and it moves down the ramp? m a) a 0 -4.9 m b) a=-4.9- m S m c) a+4.9- a, 0 2. 2 d) None of the above.

Answers

Answer:

Explanation:

When a body starts sliding on an inclined plane, the acceleration of body is due to its impotent 9f weight. The component of weight is mg Sin theta along the plane.

Thus, the acceleration is g Sin theta

= 9.8 × Sin 30 = 4.9 m/s^2

An electric coffee maker has a heating element that operates at 120 V and with a current of 2.00 A. Assuming the water absorbs all the energy delivered to the heating element, calculate the time interval (in s) during which the temperature of 0.489 kg of water rises from room temperature (23.0°C) to the boiling point. (The specific heat of water is 4,186 J/(kg · °C).)

Answers

Answer:

10.95 minute

Explanation:

V = 120 V, I = 2 A, m = 0.489 kg, c = 4186 J/kgC, T1 = 23 C , T2 = 100 C

Let time be the t.

Heat energy is equal to electrical energy

m x c (T2 - T1) = V x I x t

0.489 x 4186 x (100 - 23) = 120 x 2 x t

t = 656.73 second

t = 10.95 minute

A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of 2.20 m/s. The coefficient of kinetic friction between sled and ice is 0.115. Use energy considerations to find the distance the sled moves before it stops. m

Answers

Answer:

2.15 m

Explanation:

u = 2.2 m/s, v = 0, uk = 0.115

a = uk x g = 0.115 x 9.8 = 1.127 m/s^2

Let s be the distance traveled before stopping.

v^2 = u^2 - 2 a s

0 = 2.2^2 - 2 x 1.127 x s

s = 2.15 m

g Complete the following statements. (a) A substance that conducts electricity but whose conduction is not temperature dependent is called Correct: Your answer is correct. . (b) An example of a conductor is . (c) An example of a semiconductor is . (d) An example of an insulator is .

Answers

Answer:

Explanation:

(a) It is called conductors.

The conductors are the materials which can allow the current to pass through it.

(b) The example of conductor is copper, iron, etc.

The best conductor of electricity is silver.

(c) The example of semiconductor is silicon, germanium, etc.

The semiconductors are the materials which are insulators at normal temperature, but if the temperature increases, the conductivity of semiconductor increases.

(d) An example of insulator is wood.

Insulators are the materials which do not allow the current to pass through them.

A 200 ohm resistor has a 2-ma current in it. what is the voltage across the resistor?

Answers

Answer:

V=4v

Explanation:

To perform this operation, we will refer to Ohm's Law. This Law relates the terms current, voltage and resistance.

The current intensity that passes through a circuit is directly proportional to the voltage or voltage of the circuit and inversely proportional to the resistance it presents.

[tex]I=\frac{v}{r}[/tex]

Where

I=Current

V= Voltage

R=Resistance.

So, in this case:

R=200 ohm

I= 2mA  = 0.002 A

V= searched variable

[tex]V=r*I[/tex]

[tex]V=(200ohm)(0.002A)=4V[/tex]

Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B. B) three times smaller than the flux of electric field through Gaussian surface B. C) unrelated to the flux of electric field through Gaussian surface B. D) equal to the flux of electric field through Gaussian surface B. E) three times larger than the flux of electric field through Gaussian surface B

Answers

The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Therefore, the flux through surface A is three times larger than the flux through surface B.

The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Since the flux of electric field is proportional to the area of the surface, the flux through surface A is three times larger than the flux through surface B. Therefore, the correct answer is E) three times larger than the flux of electric field through Gaussian surface B.

Learn more about Electric flux here:

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Final answer:

The correct answer is D) equal to the flux of the electric field through Gaussian surface B. The electric flux depends on the charge enclosed by the Gaussian surface, not on its size or shape, according to Gauss's law.

Explanation:

The question asks whether Gaussian surfaces A and B, enclosing the same positive charge +Q, with surface A having three times the area of surface B, have different electric fluxes. According to Gauss's law, the electric flux (Φ) through a closed surface is proportional to the charge enclosed (Φ = Q/ε0). Since both surfaces enclose the same charge, the flux through each surface must be the same, regardless of their respective areas.

Thus, the correct answer is D) equal to the flux of the electric field through Gaussian surface B because the flux depends only on the amount of enclosed charge, not on the size or shape of the Gaussian surface.

An undiscovered planet, many light-years from Earth, has one moon which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.31500 x 10^5 km and the planet has a radius of 4.150 x10^3 km and a mass of 7.15 x10^22 kg, how long (in days) does it take the moon to make one revolution around the planet. The gravitational constant is 6.67 x10^-11 N m^2/kg^2.

Answers

Answer:

118.06 days

Explanation:

d = distance of the center of moon from surface of planet = 2.315 x 10⁵ km = 2.315 x 10⁸ m

R = radius of the planet = 4.15 x 10³ km = 4.15 x 10⁵ m

r = center to center distance between the planet and moon = R + d

M = mass of the planet = 7.15 x 10²² kg

T = Time period of revolution around the planet

Using Kepler's third law

[tex]T^{2}=\frac{4\pi ^{2}r^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4\pi ^{2}(R + d)^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4(3.14)^{2}((4.15\times 10^{5}) + (2.315\times 10^{8}))^{3}}{(6.67\times 10^{-11})(7.15\times 10^{22})}[/tex]

T = 1.02 x 10⁷ sec

we know that , 1 day = 24 h = 24 x 3600 sec = 86400 sec

T = [tex](1.02 \times 10^{7} sec)\frac{1 day}{86400 sec}[/tex]

T = 118.06 days

A closed cylindrical tank that is 8 ft in diameter and 24 ft long is completely filled with gasoline. The tank, with its long axis horizontal, is pulled by a truck along a horizontal surface. Determine the pressure difference between the ends (along the long axis of the tank) when the truck undergoes an acceleration of 5 ft/s2.

Answers

Answer:

[tex]P_{1}-P_{2}=5400lb/ft^{2}[/tex]

Explanation:

We shall use newtons second law to evaluate the pressure difference

For the system the forces that act on it as shown in the figure

Thus by Newton's second law

[tex]F_{1}-F_{2}=mass\times acceleration\\\\P_{1}\times Area-P_{2}\times Area=mass\times acceleration\\\\\because Force=Pressure\times Area\\\\\therefore P_{1}-P_{2}=\frac{mass\times acceleration}{Area}[/tex]

Mass of the gasoline can be calculated from it's density[tex]45lb/ft^{3}[/tex]

[tex]Mass=Density\times Volume\\\\Mass= 45lb/ft^{3}\times \pi \frac{d^{2}}{4}L\\\\Mass=45lb/ft^{3}\times\frac{\pi 8^{2}}{4}\times 24\\Mass=54286.72lbs[/tex]

Using the calculated values we get

[tex]P_{1}-P_{2}=\frac{54286.72\times 5}{\frac{\pi 8^{2}}{4}}[/tex]

[tex]P_{1}-P_{2}=5400lb/ft^{2}[/tex]

The total solar irradiance (TSI) at Earth orbit is 1400 watts/m2 . Assuming this value can be represented as a single Poynting flux. Find the corresponding flux at the solar visible-light surface. Explain your methods.

(Distance between Sun & Earth is 150 million km)

Answers

Answer:

3.958 × 10²⁶ watt

Explanation:

Given:

Distance between earth and sun, [tex]R_e[/tex] = 150 ×10⁸ m

Total solar irradiance at earth orbit  = 1400 watt/m²

Now,

Area irradiated ([tex]A_e[/tex]) will be = [tex]4\pi R_e^2[/tex]

⇒ [tex]A_e[/tex] =  [tex]4\pi \times (150\times 10^{8})^2[/tex]

⇒ [tex]A_e[/tex] =  [tex]2.827\times 10^{23}m^2[/tex]

Therefore, the flux = Total solar irradiance at earth orbit × [tex]A_e[/tex]

the flux =   [tex]1400watt/m^2\times 2.827\times 10^{23}m^2[/tex]

⇒the flux = 3.958 × 10²⁶ watt

At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is perpendicular to the door. What is the magnitude of the torque?

Answers

Answer:

[tex]\tau = 7.63 Nm[/tex]

Explanation:

As we know that moment of force is given as

[tex]\tau = \vec r \times \vec F[/tex]

now we have

[tex]\vec r = 1.2 m[/tex]

[tex]\vec F = 14 N[/tex]

now from above formula we have

[tex]\tau = r F sin\theta[/tex]

here we know that

[tex]\theta = 27 degree[/tex]

so we have

[tex]\tau = (1.2)(14) sin27[/tex]

[tex]\tau = 7.63 Nm[/tex]

A 15-uF capacitor is connected to a 50-V battery and becomes fullycharged. The battery is
removed and a slab of dielectric that completely fills the spacebetween the plates is inserted.
If the dielectric has a dielectric constant of 5.0:

A. what is the capacitance of the capacitor after the
slab is inserted?


B. what is the voltage across the capacitor's plates
after the slab is inserted?

Answers

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0)(15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.

What is capacitance of capacitor?

The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference.

The capacitance of the capacitor is 15-uF and it is connected to  50-V battery. The value of the dielectric constant is 5.0.

A. The capacitance of the capacitor after the slab is inserted

The dielectric constant of the slab is the ratio of capacitance of the capacitor after the slab and the capacitance of the before the slab inserted.

As the value of the dielectric constant is 5.0. Thus the capacitance of the capacitor after the slab is inserted given as,

[tex]0.5=\dfrac{C}{15\times10^{-6}}\\C=75\rm \mu F[/tex]

Thus, the capacitance of the capacitor after the slab is inserted is 75-uF.

B. The voltage across the capacitor's plates after the slab is inserted

The dielectric constant of the slab is the ratio of voltage across the capacitor's of the capacitor before the slab inserted and the voltage across the capacitor after the slab inserted.

As the value of the dielectric constant is 5.0. Thus the voltage across the capacitor's plates after the slab is inserted can be given as,

[tex]V=\dfrac{50}{5.0}\\C=10\rm V[/tex]

Thus, the voltage across the capacitor's plates after the slab is inserted is 10 volts.

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.

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The output of a generator is 440 V at 20 A. It is to be transmitted on a line with resistance of 0.60 Ω. To what voltage must the generator output be stepped up with a transformer if the power loss in transmission is not to exceed 0.010% of the original power?

Answers

Answer:

The voltage of the generator is 7.27 kV.

Explanation:

Given that,

Output of generator = 440 V

Current = 20 A

Resistance = 0.60 Ω

Power loss =0.010%

We need to calculate the total power of the generator

Using formula of power

[tex]P=VI[/tex]

Where, V = voltage

I = current

Put the value into the formula

[tex]P=440\times20[/tex]

[tex]P=8800\ W[/tex]

Th power lost on the transmission lines

[tex]P_{L}=0.010\% P[/tex]

[tex]P_{L} = 0.010\%\times8800[/tex]

[tex]P_{L}=0.88\ W[/tex]

The current passing through the transmission line

[tex]I'=\sqrt{\dfrac{P_{L}}{R}}[/tex]

[tex]I'=\sqrt{\dfrac{0.88}{0.60}}[/tex]

[tex]I'=1.211\ A[/tex]

We need to calculate the voltage of the generator

Using formula of voltage

[tex]V_{g}=\dfrac{P}{I'}[/tex]

Put the value into the formula

[tex]V_{g}=\dfrac{8800}{1.211}[/tex]

[tex]V_{g}=7.27\times10^{3}\ V[/tex]

[tex]V_{g}=7.27\ kV[/tex]

Hence, The voltage of the generator is 7.27 kV.

Final answer:

The voltage of the generator output needs to be stepped up with a transformer to limit power losses in transmission.

Explanation:

To limit power losses in transmission, the voltage of the generator output needs to be stepped up with a transformer. The formula to calculate power loss is P_loss = I^2 x R. Given that the power loss should not exceed 0.010% of the original power, we can calculate the maximum allowable power loss. By rearranging the formula, we can find the voltage required for the generator output with a transformer.

First, we calculate the original power using P = V x I. Then, we calculate the maximum allowable power loss as a percentage of the original power. Next, we substitute the given values into the formula and solve for the new voltage using the rearranged formula.

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The RC charging circuit in a camera flash unit has a voltage source of 265 V and a capacitance of 136 μF HINT (a) Find its resistance R (in ohms) if the capacitor charges to 90.0% of its final value in 16.2 s. (b) Find the average current (in A) delivered to the flash bulb if the capacitor discharges 90.0% of its full charge in 1.14 ms.

Answers

Answer:

a)

51764.7 ohm

b)

28.5 A

Explanation:

a)

t = time taken to charge to 90.0% of its final value = 16.2 s

T = time constant

C = capacitance = 136 x 10⁻⁶ F

R = resistance of the resistor

Q₀ = Maximum charge stored

Q = Charge after time "t" = 0.90 Q₀

Using the equation

[tex]Q = Q_{o}(1 - e^{\frac{-t}{T}})[/tex]

[tex]0.90 Q_{o} = Q_{o}(1 - e^{\frac{-16.2}{T}})[/tex]

[tex]0.90 = (1 - e^{\frac{-16.2}{T}})[/tex]

T = 7.04 s

Time constant is given as

T = RC

7.04 = R (136 x 10⁻⁶)

R = 51764.7 ohm

b)

V = Potential difference of voltage source = 265 volts

q = Amount of charge discharged = 0.90 Q₀ = 0.90 (CV) = (0.90) (136 x 10⁻⁶) (265) = 0.032436 C

t = time taken to discharge = 1.14 x 10⁻³ s

Current is given as

[tex]i = \frac{q}{t}[/tex]

i = 28.5 A

For the RC charging circuit of a camera flash unit has the value of resistance and current as,

(a)The resistance R is 51765 ohms.(b) The average current  delivered to the flash bulb is 28.5 ampere.

What is the formula of RC circuit?

RC circuit is the circuit in which the voltage or current passes through the resistor and capacitor consist by the circuit.

The formula for the RC circuit can be given as,

[tex]Q=Q_o\left(1-e^{\dfrac{-1}{RC}}\right )[/tex]

Here, [tex](Q_o)[/tex] is the initial charge and (R) is the resistance.

(a)The resistance R-

The  capacitor charges to 90.0% of its final value. Thus, the charge for the first case can be given as,

[tex]Q=0.90 Q_o[/tex]

The capacitance of 136 μF. Put the values in the above formula as,

[tex]0.90Q_o=Q_o\left(1-e^{\dfrac{-1}{R\times136\times10^{-6}}}\right )\\R=51765\rm ohm[/tex]

Thus, the value of the resistance R is 51765 ohms.

(b) The average current  delivered to the flash bulb-

The voltage source of the RC circuit is 265 V  and the capacitor discharges 90.0% of its full charge in 1.14 ms. Thus, the amount of charge discharged is,

[tex]Q=0.9\times136\times10^{-6}\times265\\Q=3.2436\times10^{-2}\rm C[/tex]

The value of current is the ratio of discharge per second time (1.14 ms.). Therefore,

[tex]I=\dfrac{3.2436\times10^{-2}}{1.14\times10^{-3}}\\I=28.5\rm A[/tex]

Thus, for the RC charging circuit of a camera flash unit has the value of resistance and current as,

(a)The resistance R is 51765 ohms.(b) The average current  delivered to the flash bulb is 28.5 ampere.

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A rhinoceros beetle rides the rín of a small disk that rotates like a merry-go-round. If the beetle crawls toward the center of the disk, do the following (each relative to the central axis) increase, decrease, or remain the same for the beetle-disk system: (a) rotational inertia, (b) angular momentum, and (c) angular speed?

Answers

Answer:

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

Explanation:

Here the Beetle is initially moving along the rim of the disc

So here during the motion of beetle there is no external force on the system of beetle and the disc.

So here we can also say that there is no torque acting on the system

so angular momentum of the disc + beetle system will remain conserved

so here we have

[tex]I_1\omega_1 = I_2 \omega_2[/tex]

here as the beetle crawls towards the centre of the disk then

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

A string of length L= 1.2 m and mass m = 20g is under 400 N of tension, its two ends are fixed. a. How many nodes will you see in the 5th harmonic.
i. 4
ii. 5
iii. 6
iv. 7
v. None of the above.

Answers

Answer:

(iii) 6

Explanation:

Part a)

Since it is given that both ends are fixed and it is vibrating in 5th harmonic

So here it will have 5 number of loops in it

so we can draw it in following way

each loop will have 1 antinode and two nodes which means number of nodes is one more than number of anti-nodes

So there are 5 loops which means it will have 5 antinodes

and hence there will be 6 nodes in it

so correct answer will be

(iii) 6

A bicycle with 0.80-m-diameter tires is coasting on a level road at 5.6 m/s . A small blue dot has been painted on the tread of the rear tire. Part A What is the angular speed of the tires? Express your answer in radians per second. ω ω = 14 rad/s Previous Answers Correct Part B What is the speed of the blue dot when it is 0.80 m above the road?

Answers

Answer:

a)

14 rad/s

b)

11.2 m/s

Explanation:

a)

d = diameter of tire = 0.80 m

r = radius of tire = (0.5) d = (0.5) (0.80) = 0.40 m

v = speed of bicycle = 5.6 m/s

w = angular speed of the tire

Speed of cycle is given as

v = r w

5.6 = (0.40) w

w = 14 rad/s

b)

v' = speed of blue dot

Speed blue of dot is given as

v' = v + rw

v' = 5.6 + (0.40) (14)

v' = 11.2 m/s

Final answer:

The angular speed of the bicycle tire is 14 rad/s. The speed of the blue dot when it is 0.80 m above the road is the same as the bicycle's speed which is 5.6 m/s.

Explanation:

This is a problem which involves the calculation  of the angular speed and linear speed of an object in circular motion, here the circular motion being the rotation of a bicycle tire.

Given the linear speed of the bicycle and the radius of the tire, we would use the equation that links linear speed v and angular speed a, expressed as v = ra. We can find angular speed by rearranging this formula as a = v/r. Hence, we can calculate the angular speed of the tire as 5.6 m/s divided by the radius of the tire, which is half of the diameter, so 0.4m, which equals 14 rad/s.

Next, when a point on the tire (the blue dot) is 0.8 m above the ground it is at the top of tire's circular path, so its speed is equivalent to the linear speed of the bicycle, 5.6 m/s. This is because at this point the dot is not in contact with the ground hence it isn't stationary relative to the ground unlike the point at the bottom of the tire.

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The 400 kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F= (3200) N, where t is in seconds. If the car has an initial velocity V1= 2m/s at s 0 and t= 0, determine the distance it moves the plane when (a) t 1 s and (b) f-5 s.

Answers

Answer:

(a) 110 m/s

(b) 42 m/s

Explanation:

mass, m = 400 kg, F = 3200 N, V1 = 2 m/s,

acceleration, a = Force / mass = 3200 / 400 = 8 m/s^2

(a) Use first equation of motion

v = V1 + a t

v = 2 + 8 x 1 = 10 m/s

(b) Again using first equation of motion

v = V1 + a t

v = 2 + 8 x 5 = 42 m/s

Thus, the velocity of plane after 1 second is 10 m/s and after 5 second the velocity is 42 m/s.

Final answer:

The 400 kg mine car moves a distance of 6 m after one second and 110 m after five seconds given an initial velocity of 2 m/s and a force of 3200 N.

Explanation:

The problem describes a physics scenario where a 400 kg mine car is pulled up an incline by a cable and motor. The force on the cable is given as F = 3200N and the initial velocity, V1, is given as 2 m/s. We can calculate the distance the car moves on the plane at different times using the physics equations of motion.

Let's use the equation of motion: s = ut + 1/2 at², where 's' is the distance moved, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.

Given that the net force is equal to mass times acceleration (F = ma), we can calculate the acceleration, 'a', as F/m. So, a = 3200N/400kg = 8 m/s².

(a) t = 1s: The distance moved is s = 2m/s * 1s + 1/2 * 8 m/s² * (1s)² = 2m + 4m = 6m.(b) t = 5s: The distance moved is s = 2m/s * 5s + 1/2 * 8 m/s² * (5s)² = 10m + 100m = 110m.

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Identify the least dense and most dense planets and provide an explanation as to why they are the least and most dense.

Answers

Answer:

Explained

Explanation:

The most dense planet in  our solar system is Earth. Earth is most dense planet because

Earth is made of dense material like rocks, core is made of metals, soil and water. The average density of earth is 5.5 gram/cm^3.The overall mass of Earth is higher than those of other rocky planets. This heaviness accounts for higher density of earth.

The least dense planet of the solar system is Saturn because Saturn is mostly made of gases and its size is smaller than the Jupiter. Jupiter has more gravity hence its density is higher to Saturn. Moreover, Uranus and Neptune are ice giants. Although they are also made of gases but due their distance from sun most of these gases have solidified. Hence making them more dense than Saturn and Jupiter.

Final answer:

Saturn is the least dense planet, while Earth is the densest planet in our solar system.

Explanation:

The least dense planet in our solar system is Saturn, while the densest planet is Earth. Saturn is a gas giant composed mostly of hydrogen and helium, which have lower densities compared to the rocks and metals found on rocky planets like Earth. Earth, on the other hand, has a higher density due to its solid composition and a core made of iron and nickel.

A bolt comes loose from underneath an elevator that is moving upward at a speed of 5 m/s. The bolt reaches the bottom of the elevator shaft in 3.1 s. (a) How high up was the elevator when the bolt came loose? (In m)
(b) What is the speed of the bolt when it hits the bottom of the shaft? (In m/s)

Answers

Answer:

a)  The elevator was  31.64 m high up when the bolt came loose.

b)  Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

Explanation:

a) Considering motion of bolt:-

Initial velocity, u =  5 m/s

Acceleration , a = -9.81 m/s²

Time = 3.1 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= 5 x 3.1 - 0.5 x 9.81 x 3.1²

    s = 0 x t + 0.5 x 9.81 x t²

    s = -31.64 m

The elevator was  31.64 m high up when the bolt came loose.

b) We have equation of motion v = u + at

  Initial velocity, u =  5 m/s

 Acceleration , a = -9.81 m/s²

 Time = 3.1 s  

Substituting

  v = u + at

  v  = 5 - 9.81 x 3.1 = -25.41 m/s

Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

How fast must a meter stick be moving if its length is observed to shrink to 0.7 m?

Answers

Answer:

2.14×10⁸ m/s

Explanation:

L=Required Length=0.7 m

L₀=Initial length=1 m

c=speed of light=3×10⁸ m/s

[tex]From\ Lorentz\ Contraction\ relation\\L=L_0\sqrt{1-\frac {v^2}{c^2}}\\\Rightarrow 0.7=1\sqrt {1-\frac {v^2}{c^2}}\\\Rightarrow 0.49=1-\frac {v^2}{c^2}\\\Rightarrow 0.49-1=-\frac {v^2}{c^2}\\\Rightarrow -0.51=-\frac {v^2}{c^2}\\\Rightarrow 0.51=\frac {v^2}{c^2}\\\Rightarrow v^2=0.51\times c^2\\\Rightarrow v=\sqrt{0.51} \times c\\\Rightarrow v=0.71\times 3\times 10^8\\\Rightarrow v=2.14\times 10^8\ m/s\\\therefore velocity\ of\ stick\ should\ be\ 2.14\times 10^8\ m/s[/tex]

A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time interval?

Answers

Answer:

Number of electrons, [tex]n=5.62\times 10^{21}[/tex]

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, [tex]I=\dfrac{q}{t}[/tex]

[tex]I\times t=n\times e[/tex]

[tex]n=\dfrac{It}{e}[/tex]

e is the charge of an electron

[tex]n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}[/tex]

[tex]n=5.62\times 10^{21}[/tex]

So, the number of electrons pass through the resistor is [tex]5.62\times 10^{21}[/tex]. Hence, this is the required solution.

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