A resistor with R=300 ? and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 216 W. (a) What is the impedance Z of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?

Answers

Answer 1

Answer:

impedance Z = 416.66 ohm

voltage across inducance V = 346.99 V

power factor = 0.720

Explanation:

given data

resistor R = 300

voltage amplitude = 500 V

resistor = 216 W

to find out

impedance and amplitude of the voltage and power factor

solution

we apply here average power formula that is

average power = I²×R     ............1

I = [tex]\frac{Vrms}{Z}[/tex]

so

average power =  ([tex]\frac{Vrms}{Z}[/tex])²×R

Vrms = [tex]\frac{1}{\sqrt{2} }[/tex] × Vmax

Z = V × [tex]\sqrt{\frac{R}{2P} }[/tex]

Z = 500 × [tex]\sqrt{\frac{300}{2*216} }[/tex]

impedance Z = 416.66 ohm

and

we know voltage across inductor is here express as

V = I × X     .............2

so here X will be by inductance

Z² = R² + (X)²  

(X)²  = 416.66² - 300²  

X = 289.15 ohm

and I = [tex]\frac{V}{Z}[/tex]

I = [tex]\frac{500}{416.66}[/tex]

I = 1.20 A

so from equation 2

V = 1.20 × 289.15

voltage across inducance V = 346.99 V

and

average power = Vmax × Imax  ×  cos∅

tan∅ = [tex]\frac{289.15}{300}[/tex]

tan∅ = 43.95°

so power factor is

power factor = cos43.95°

power factor = 0.720

Answer 2
Final answer:

The problem involves the use of Ohm's law, impedance, power, and a power factor in an AC circuit. The impedance of the circuit, the voltage across the inductor, and the power factor can be calculated using given values, the principle of impedance and the relationships among AC voltage, current, resistance and power.

Explanation:

The question involves an AC circuit composed of a resistor and inductor in series connected to an AC voltage source. We have a resistor with a resistance (R) of 300 Ω and a dissipated power (P) of 216W. The voltage amplitude (Vo) of the AC source is 500 V. It is important to remember that in this context, impedance (Z), which has a unit of ohms, represents the total resistance in an AC circuit and can be calculated using Ohm's law for AC circuits.

(a) To find the impedance Z of the circuit, we consider that the power P is given by the relation P = Vo^2/R, substituting for P, and R, we can solve for Vo, which will be sqrt(P*R). Then, the rms voltage (V) is given by Vo/sqrt(2). Our current I would be P/V. Finally, applying Ohm's law, Z=V/I would give us the impedance.

(b) The voltage across the inductor can be found by using Pythagoras' Theorem in the context of an AC circuit, VL = sqrt(Vo^2 - VR^2), where VR is the voltage across the resistor (equal to I* R).

(c) Lastly, the power factor can be found as the cosine of the phase angle θ, which can also be defined as R/Z. We'd first calculate θ = arccos(R/Z), and then find the power factor as cos(θ).

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Related Questions

10 kg of liquid water is in a container maintained at atmospheric pressure, 101325 Pa. The water is initially at 373.15 K, the boiling point at that pressure.The latent heat of water -> water vapor is 2230 J/g. The molecular weight of water is 18 g.103 J of heat is added to the water.1)How much of the water turns to vapor?mass(vapor)=

Answers

Answer:

[tex]m=0.0462\ g[/tex] of water is converted into vapour.

Explanation:

Given:

mass of water, [tex]m_w=10\ kg[/tex]pressure conditions, [tex]P=101325\ Pa[/tex]temperature conditions, [tex]T=373.15\ K[/tex]latent heat of vapourization of water, [tex]L=2230\ J.g^{-1}[/tex]amount of heat supplied to the water, [tex]103\ J[/tex]

Now using the equation of heat considering latent heat only:

(since water already at boiling point at atmospheric temperature)

[tex]Q=m.L[/tex]

[tex]103=m\times 2230[/tex]

[tex]m=0.0462\ g[/tex] of water is converted into vapour.

Final answer:

In the situation provided, about 46.2 grams of the water will have transitioned from a liquid to a vapor after being supplied with 103 kJ of energy, given the specified latent heat of vaporization.

Explanation:

Given that the latent heat of water's vaporization is 2230 J/g and 103 J of energy was provided to the water, we first convert all our units to be consistent. Remember that the latent heat of vaporization is the amount of heat energy required to change one gram of a substetance from a liquid to a gas at constant mperature and pressure. In this case, we're transitioning water to water vapor.

The input energy is 103 kJ, and the latent heat of vaporization is 2.23 kJ/g, so we can calculate the mass of the water that was vaporized using the equation: mass (g) = energy input (kJ) / latent heat of vaporization (kJ/g). By plugging in the values we get: mass = 103 / 2.23 = 46.2 grams.

So, approximately 46.2 grams of the water will have transitioned from a liquid to a vapor given the provided energy input of 103 kJ.

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Recall that force is a change in momentum over a change in time, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiative momentum striking the sail per second. What is the approximate magnitude of the pressure on the sail in the vicinity of Earth’s Orbit?

Answers

Answer:

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= [tex]\frac{2I}{c}[/tex]

Explanation:

The momentum of a photon is:

p = E/c

E = the photon energy

c = the speed of light.

take the time derivative (gives the force)

F = dp/dt = (dE/dt)/c

F = 2(dE/dt)/c (is doubled for complete reflection of the light)

Intensity has the units of energy per unit time per unit area

=  I

then,

Force/unit area = 2I/c

magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= [tex]\frac{2I}{c}[/tex]

Select the correct statement to describe when a sample of liquid water vaporizes into water vapor

Answers

Answer:

This procces is called evaporation.

Explanation:

When you have liquid water that is transformed into steam, a phase change is called evaporation. The temperature for the evaporation of water depends on the pressure, for example for water at atmospheric pressure the temperature of evaporation is equal to 100°C. as the pressure increases are achieved evaporation temperatures higher. When that happens, the phase change temperature of the water is not increasing, as the process that takes place is the transfer of latent heat and applies only to changes of phase, that is to say at atmospheric pressure when it has 100% of the steam this will be at 101°C.

A container made of steel, which has a coefficient of linear expansion 11 ✕ 10−6 (°C)−1, has a volume of 55.0 gallons. The container is filled to the top with turpentine, which has a coefficient of volume expansion of 9.0 ✕ 10−4 (°C)−1, when the temperature is 10.0°C. If the temperature rises to 25.5°C, how much turpentine (in gal) will spill over the edge of the container?

Answers

Final answer:

The amount of spilled turpentine due to temperature rise can be calculated by comparing the changes in volume of the steel container and the contained turpentine, using the principles and mathematical formulas of thermal expansion.

Explanation:

The question asks us to determine how much turpentine, which exhibits a greater rate of thermal expansion compared to the steel container, will spill over when the temperature rises. We need to apply the principles of thermal expansion to calculate this. Both the steel container and the turpentine expand with the increase in temperature, but since turpentine has a higher coefficient of volume expansion than steel, more turpentine will expand than the container can accommodate, resulting in some turpentine spilling over.

To calculate the amount of spilled turpentine, we need to find the change in volume for both the container and turpentine, and subtract the former from the latter. The change in volume due to thermal expansion can be calculated by using the equation ΔV = βV0*(T2 - T1), where β is the coefficient of volume expansion, V0 is the initial volume, and T2 and T1 are the final and initial temperatures respectively.

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Final answer:

To find how much turpentine will overflow, calculate the change in volume due to thermal expansion for both the steel container and the turpentine, and subtract the change in volume of the steel from that of the turpentine.

Explanation:

To solve this problem, we need to consider the thermal expansion of both the steel container and the turpentine. Thermal expansion is the increase in size of a body due to a change in temperature. This is described mathematically by the formula ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the original volume, and ΔT is the change in temperature.

First, calculate the change in volume for the steel container and the turpentine separately. For the steel, β is 11 x 10^-6 (°C)^-1 and for turpentine, β is 9.0 x 10^-4 (°C)^-1. The original volume, V₀, is 55 gallons for both, and the change in temperature, ΔT, is 25.5°C - 10.0°C = 15.5°C.

Performing these calculations will give you the change in volume for the steel and the turpentine. The difference in these two volumes will tell you how much turpentine will overflow as the temperature increases.

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A 30 gram bullet is shot upward at a wooden block. The bullet is launched at the speed vi. It travels up 0.40 m to strike the wooden block. The wooden block is 20 cm wide and 10 cm high and its thickness gives it a mass of 500 g. The center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height. a. What is the launch speed of the bullet? b. How much mechanical energy does the bullet and the block system have before all of the processes? Use the surface the block rests on as the reference for where gravitational potential energy is zero. c. How much mechanical energy does the bullet and the block system have after all of the processes? d. How much mechanical energy was lost from beginning to end?

Answers

Answer:

Explanation:

Mass of bullet m = .03 kg

Mass of wooden block M = 0.5 kg

Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height

Velocity of wooden block + bullet just after impact = √2gH

=√(2 x 9.8 x 0.6)

= 3.43 m / s

Let the launch velocity of bullet be v₁

If v₂ be the velocity with which bullet hits the block

Applying law of conservation of momentum

.03 x v₂ = .530 x 3.43

v₂ = 60.6 m /s

if v₁ be initial velocity

v₂² = v₁² - 2 gh

v₁² = v₂² + 2 gh

= 60.6 ² + 2 x 9.8 x 0.4

v₁ = 60.65 m /s this is launch speed.

b )

Initial kinetic energy of bullet

= 1/2 m v²

= .5 x .03 x 3680

= 55 J

Potential energy of bullet + block = 0

Total energy = 5 J

c)

Kinetic energy of bullet block system

1/2 m v²

= .5 x .53 x  3.43

= 3.11 J

d )

Loss of energy in the impact =  Total mechanical energy  lost from beginning to end?

3.11 J  - 5

= 1.89 J

A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 75°C. The other sample has a mass of 66 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific heat of the unknown second sample. (Pick the answer closest to the true value.)A. 1950 joules Co/kgB. 975 joules Co/kgC. 3950 joules Co/kgD. 250 joules Co/kgE. 8500 joules Co/kg

Answers

Answer:

A. 1,950 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC  = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC

Final answer:

The specific heat of the unknown metal can be determined from the equilibrium of heat transfer in the system. The heat lost by the hot substances is equal to the heat gained by the cooler substances. Solving for the specific heat of the unknown substance involves calculating the heat gained and lost and equating their values.

Explanation:

The specific heat of a substance is a measure of the amount of heat energy required to raise the temperature of a certain mass of the substance by a certain amount. In this case, we're solving for the specific heat (c) of an unknown substance. As the system is in thermal equilibrium, the heat lost by hot substances (copper and unknown metal) is equal to the heat gained by the cooler substances (water and the calorimeter).

The specific heat (c) of the unknown substance can therefore be determined by setting the heat gained (Q_gained = m*c*ΔT) by the cooler substances equal to the heat lost (Q_lost = m*c*ΔT) by the hot substances and solving for the specific heat (c) of the unknown substance. Given that ΔT is the change in temperature, m is the mass, and c is the specific heat, and using the specific heat values for water, aluminum, and copper.

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Car drag racing takes place over a distance of a mile (402 m) from a standing start. If a car (mass 1600 kg) could be propelled forward with a pulling force equal to that of gravity, what would be the change in kinetic energy and the terminal speed of the car (in mph) at the end of the race be? (For comparison, a modern, high-performance sports car may reach a terminal speed of just over 100 mph = 44.7 m/s.)

Answers

Answer:

v = 88.76 m / s ,  K = 6.30 10⁶ J

Explanation:

For this exercise the force that is applied is that necessary for the acceleration of the car to be the acceleration of gravity, they do not indicate that there is friction, we look for the final speed

       v² = v₀² + 2 a x

Since the car starts from rest, the initial speed is zero, vo = 0

       v = √ 2 a x

       v = √ (2 9.8 402)

       v = 88.76 m / s

Let's look for kinetic energy

       K = ½ m v²

       K = ½ 160kg 88.76²

       K = 6.30 10⁶ J

Why is it impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally?A. In an orbiting station, the gravitational force is too large and the astronaut can't take his feet off the floor.B. It is impossible to walk inside an orbiting space station because its rotation is too fast.C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut "jumps" on the same place.D. In an orbiting station, after one foot pushes off there isn't a force to bring the astronaut back to the "floor" for the next step.

Answers

Final answer:

An astronaut cannot walk normally in a space station because there's no frictional force to move forward in the near-weightless environment. To move, astronauts use handholds and walls, pushing against them to create a reaction force.

Explanation:

It is impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally because C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut would indeed "jump" in place due to the lack of friction between their feet and the floor of the space station, which is a result of the near-weightlessness they experience. In space, normal walking is ineffective because walking relies on gravity to pull the body back down to the floor after each step, which isn't present in the same way on a space station in orbit.

In order to move in such an environment, an astronaut must push against a solid object, creating a reaction force in the opposite direction according to Newton's third law of motion. This principle allows the astronaut to propel and steer themselves around the space station using handholds and walls. The environment inside the ISS is similar to that inside a freely falling box where gravity still exists, but occupants do not feel its effects because they are in free fall around Earth, which creates the sensation of weightlessness.

Final answer:

Astronauts cannot walk normally in an orbiting space station due to the lack of gravity and friction. They are in a state of free fall, creating a sensation of weightlessness. Movement can be achieved by utilizing the conservation of momentum and Newton's third law of motion. Therefore option C is the correct answer.

Explanation:

The reason it is impossible for an astronaut inside an orbiting space station to walk from one end to the other by walking normally is C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut cannot walk from one end to the other by walking normally because, in the microgravity environment of an orbiting spacecraft, traditional walking, which relies on the force of gravity and friction between the feet and the ground, does not work. Instead, astronauts move about by pushing off surfaces or floating through the air.

In orbit, the International Space Station (ISS) and everything inside it, including the astronauts, are in a state of free fall. They are falling around Earth at the same rate as the space station, creating a sensation of weightlessness. This is akin to the sensation of temporary weightlessness one experiences at the topmost point of a roller coaster ride or when an elevator suddenly descends.

Achieving locomotion for an astronaut stranded in the center of the station without contact with any solid surface would necessitate a method that does not rely on gravity or friction. The astronaut would have to utilize the principle of conservation of momentum. For instance, by throwing an object in one direction, the astronaut would move in the opposite direction, as described by Newton's third law of motion: for every action, there is an equal and opposite reaction.

A rigid, insulated tank that is initially evacuated is connected though a valve to a supply line that carries steam at 1 MPa and 300∘C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed.

Determine the final temperature of the steam in the tank, in ∘C.

Answers

Answer:

Final Temperature of the steam tank = 456.4°C

Explanation:

Assuming it to be a uniform flow process, kinetic and potential energy to be zero, and work done and heat input to be zero also.  We can conclude that,

Enthalpy of the steam in pipe = Internal Energy of the steam in tank

Using the Property tables and Charts - Steam tables,  

At Pressure= 1 MPa and Temperature= 300°C,

Enthalpy = 3051.2 kJ/kg

At Pressure= 1 MPa and Internal Energy= 3051.2 kJ/kg,

Temperature = 456.4°C.

A navy seal of mass 80 kg parachuted into an enemy harbor. At one point while he was falling, the resistive force of air exerted on him was 520 N. What can you determine about the motion?

Answers

Answer:

The motion of the parachute = 3.3 m/s²

Explanation:

Weight of the parachute - Resistive force of air = ma

W - Fₐ  = ma.................... Equation 1

making a the subject of formula in equation 1

a = (W- Fₐ)/m.................. Equation 2

Where W = weight of the parachute, Fₐ = resistive force of air, m = mass of the parachute, a = acceleration of the parachute

Constant: g = 9.8 m/s²

Given: Fₐ = 520 N, m = 80 kg

W = mg = 80 × 9.8 = 784 N,

Substituting these values into equation 2

a = (784-520)/80

a = 264/80

a = 3.3 m/s²

Therefore the motion of the parachute = 3.3 m/s²

In the fastest measured tennis serve, the ball left the racquet at 73.14 m/s. A serve tennis ball is typically in contact with the racquet for 30.0 ms and starts from rest. Assume constant acceleration.(a) what was the ball's acceleration during this serve??(b) how far did the ball travel during the serve???

Answers

Answer:

a)  the acceleration is a= 2438 m/s²

b) the distance travelled during serve is d = 1.0971 m

Explanation:

a) since

v = vo + a*t ,

where v= velocity at time t , vo= velocity at time t=0 and a= acceleration

,then

a= (v-vo)/t

replacing values

a= (v-vo)/t = (73.14 m/s - 0 m/s)/( 30* 10⁻³ s) = 2438 m/s²

b) the distance travelled d is

v² = vo² + 2*a*d  

then

d = (v² - vo²) /(2*a) = (73.14 m/s)² - 0²)/(2*2438 m/s²)= 1.0971 m

a)  the acceleration is a= 2438 m/s²

b) the distance travelled during serve is d = 1.0971 m

What is acceleration?

Acceleration represents the rate at which velocity should be changed with time, with respect to both speed and direction. Since acceleration contains both a magnitude and a direction, it is a vector quantity.

Calculation of acceleration & distance:

a) since

[tex]v = vo + a\times t[/tex]

Here

v= velocity at time t ,

vo= velocity at time t=0

and a= acceleration

Now

[tex]a= (v-vo)\div t\\\\ =(73.14 m/s - 0 m/s)/( 30\times 10^{-3} s)[/tex]

= 2438 m/s²

b) Now the distance traveled d is

[tex]v^2 = vo^2 + 2\times a\times d \\\\d = (v^2 - vo^2) \div (2\timesa) \\\\=(73.14 m/s)^2 - 0^2)\div (2\times 2438 m/s^2)[/tex]

= 1.0971 m

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8–4. The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.

Answers

Answer:

The stress S = 1935 [Psi]

Explanation:

This kind of problem belongs to the mechanical of materials field in the branch of the mechanical engineering.

The initial data:

P = internal pressure [Psi] = 90 [Psi]

Di= internal diameter [in] = 22 [in]

t = wall thickness [in] = 0.25 [in]

S = stress = [Psi]

Therefore

ri = internal radius = (Di)/2 - t = (22/2) - 0.25 = 10.75 [in]

And using the expression to find the stress:

[tex]S=\frac{P*D_{i} }{2*t} \\replacing:\\S=\frac{90*10.75 }{2*0.25} \\S=1935[Psi][/tex]

In the attached image we can see the stress σ1 & σ2 = S acting over the point A.

Did you think about this over Christmas? I did (-: Before Christmas a 65kg student consumes 2500 Cal each day and stays at the same weight. For three days in a row while visiting her parents she eats 3500 Cal and, wanting to keep from gaining weight decides to "work off" the excess by jumping up and down at the Christmas tree. With each jump she accelerates to a speed of 3.2 m/s before leaving the ground. a) How high will she jump each jump? b) How many jumps must she do to keep her weight? Assume that the efficiency of the body in using energy is 25%. c) Do you suggest that is a reasonable way for the student not to gain weight over Christmas? d) Possible enhancement: What other way/ways would you suggest for the student to keep her weight?

Answers

Answer:

a)  Em = 332.8 J , b) # jump = 13, c)   It is reasonable since there are not too many jumps , d) lower the calories consumed

Explanation:

a) Let's use energy conservation

Initial. On the floor

             Em₀ = K = ½ m v²

Final. The highest point

             Emf = U = m g h

Energy is conserved

             Em₀ = Emf

             ½ m v² = m g h

             h = ½ v² / g

            h = ½ 3.2² /9.8

            h = 0.52 m

b) When he was at home he maintained his weight with 2500 cal / day. In his parents' house he consumes 3500 cal / day, the excess of calories is

            Q = 3500 -2500 = 1000cal / day

Let's reduce this value to the SI system

             Q = 1000 cal (4,184 J / 1 cal) = 4186 J / day

Now the energy in each jump is

               Em = K = ½ m v²

               Em = ½ 65 3.2²

               Em = 332.8 J

They indicate that the body can only use 25% of this energy

              Em effec = 0.25 332.8 J

              Em effec = 83.2 J

This is the energy that burns the body

Let's use a Proportion Rule (rule of three), if a jump spends 83.2J how much jump it needs to spend 1046 J

              # jump = 1046 J (1 jump / 83.2 J)

              # jump = 12.6 jumps / day

              # jump = 13  

c) It is reasonable since there are not too many jumps

d) That some days consume more vegetables to lower the calories consumed

An observer sits in a boat watching wave fronts move past the boat. The distance between successive wave crests is 0.80 m, and they are moving at 2.2 m / s.

What is the wavelength of these waves?
a. 1.6 m
b. 2.2 m
c. 0.80 m

What is the frequency of these waves?
a. 0.36 Hz
b. 2.8 Hz
c. 0.80 Hz

What is the period of these waves?
a. 0.80 s
b. 0.36 s
c. 2.8 s

Answers

To solve this problem we will use the three requested concepts: Wavelength, frequency and period.

The wavelength is the distance between each crest, therefore it is already given and is 0.8m

The correct answer is C.

The frequency can be described as a relationship between wave speed and wavelength therefore

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

[tex]f = \frac{2.2}{0.8}[/tex]

[tex]f = 2.75Hz \approx 2.8Hz[/tex]

The correct answer is B.

The period is the inverse of the frequency therefore

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{2.8}[/tex]

[tex]T = 0.35s[/tex]

The correct answer is B.

(a) The wavelength of the wave is 0.80m and the right option is c.

(b) The frequency of the wave is 2.8 Hz and the right option is b.

(c) The period of the wave is 0.36 s and the right option is b

(a) The distance between successive wave crests = wavelength of the wave

From the question,

(a) Wavelength = 0.80 m

Hence the wavelength = 0.80 m

(b) Using,

     V = λf.............. Equation 1

Where V = Velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

f = V/λ.................... Equation 2

Given: V = 2.2 m/s, λ = 0.80 m

Substitute these values into equation 2

f = 2.2/0.8

f = 2.75 Hz.

f ≈ 2.8 Hz

Hence the frequency of the wave is 2.8 Hz

(c) f = 1/T.............. Equation 3

Where T = period.

Therefore,

T = 1/f .................. 4

Given: f = 2.8 Hz,

T = 1/2.8

T = 0.357

T ≈ 0.36 s

Hence the period of the wave = 0.36 s

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You are playing a speed-based card game with your 64-year-old grandfather. The object of the game is to get rid of your cards as fast as you can. Once the first card is turned over, each player tries to play by deciding on which pile to play his or her card. When you were younger, your grandfather always beat you in this game. Now, you always beat him. Your grandfather is likely experiencing a slight decline in_____________.

Answers

Now, you always beat him. Your grandfather is likely experiencing a slight decline in perceptual speed.

Explanation:

The speed of perception refers to the capacity to accurately (and completely) compare words letter, digits, objects, images, etc. When testing, these objects can be displayed simultaneously or one after the other. This type of test can be included in the proficiency test.

For example, we have also seen all the puzzles that ask the reader to notice the differences between the two pictures. The time it takes to recognize these differences is a measure of the speed of perception. Likewise, in getting rid of cards at the given situation, grandfather experiences a less decline in his perceptual speed.

Your grandfather is likely experiencing a slight decline in; Perceptual speed.

The grandfather is playing a speed based card game.

Now we are told that the object of the game is to get rid of the cards as fast as possible.

We are told that when you were younger your grandfather used to beat you always in the game. This means that his speed in comparing the cards to know which one to get rid of was fast before but has declined now since he can't beat you again.

Finally, we can say that his perceptual speed has declined because perceptual speed is defined as the ability to compare letters, numbers, objects, patterns e.t.c

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Spaceships A and B are traveling directly toward each other at a speed 0.5c relative to the Earth, and each has a headlight aimed toward the other ship.


What value do technicians on ship B get by measuring the speed of the light emitted by ship A's headlight?


Answer choices


1).75c


2)1.0c


3)1.5c


4) .5c

Answers

The answer is 2) 1.0c. Light will always propagate through a vacuum at the speed of light “c”; even when moving at a significant fraction of the speed of light, observers will still measure this as the speed of light and the difference is resultant of time dilation.

When he prepares food, Edgar wants to use ingredients that are not high in trans fat. Based on this information, which of the following fats should he include in his recipes?a. Stick margarine made with olive oilb. Partially-hydrogenated peanut oilc. Corn oild. Shortening

Answers

Answer:

Corn oil

Explanation:

Researchers say that corn oil is safer option than olive oil when it comes to reducing sugar levels of the blood and cholesterol levels and that it is often successful in reducing blood pressure.  Corn oil is a polyunsaturated fat with some monounsaturated properties. Also, Corn oil is not high in trans fat responsible for various diseases.

Because sunspots are a little cooler than the average temperature of the photosphere, they prevent some energy from being released from the part of the surface they occupy, but this energy is usually released from hotter and brighter than average areas nearby. Please answer the question: What could happen to the Sun if this energy release did not happen

Answers

Answer: Corona Mass Ejections(CME) and Solar flares

Explanation: Corona Mass Ejections and Solar flares are eruptions that occur in the sun due to the instability in the magnetic field of the sun. This Corona Mass Ejections and Solar flares are prevented by sun spots.Corona mas Ejections are Large and massive eruptions,solar flares are somewhat small eruptions,both of them are prevented from occuring by Sunspots which help to dissipate cooler temperatures in the sun.

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

(A) The magnitude of the tension increases to four times its original value, 4F.
(B) The magnitude of the tension reduces to half of its original value, F/2.
(C) The magnitude of the tension is unchanged.
(D) The magnitude of the tension reduces to one-fourth of its original value, F/4.
(E) The magnitude of the tension increases to twice its original value, 2F.

Answers

Final answer:

When a stone is whirled at double the speed, the tension in the string increases to four times its original value, assuming the radius of the whirl remains the same.

Explanation:

The tension in a string whirling a stone in a circle at a constant speed is directly proportional to the square of the speed. If the boy doubles the speed in the scenario you gave, keeping the radius of the circle unchanged, the tension in the string would increase as the square of that factor. So, between the options given, if the boy increases the speed of the stone so that it makes two complete revolutions every second instead of one, the magnitude of the tension in the string increases to four times its original value. Thus, the correct answer is (A) the magnitude of the tension increases to four times its original value, 4F.

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Final answer:

The tension in the string of a whirling stone increases by a factor of four when the speed of rotation doubles and the radius remains the same. It is because the tension is directly proportional to the square of the speed of the stone.

Explanation:

The tension in the string of a whirling stone is related to the centripetal force, which is directly proportional to the square of the speed of rotation and the mass of the stone, and inversely proportional to the radius of the circle. If the speed of rotation doubles (from one revolution per second to two revolutions per second) and the radius of the circle remains the same, the resulting tension in the string (centripetal force) increases by a factor of four.

Hence, the answer is (A) The magnitude of the tension increases to four times its original value, 4F.

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A forward-biased silicon diode is connected to a 12.0-V battery through a resistor. If the current is 12 mA and the diode potential difference is 0.70 V, what is the resistance?

Answers

To solve this problem we will use the concepts related to Ohm's law for which voltage, intensity and resistance are related.

Mathematically this relationship is given as

[tex]V = IR \rightarrow R= \frac{V}{I}[/tex]

Where,

V= Voltage

I = Current

R = Resistance

The value of the given voltage is 12V, while the current is 12mA, therefore the resistance would be

[tex]R = \frac{12}{12*10^{-3}}[/tex]

[tex]R = 1000 \Omega[/tex]

Therefore the resistance is [tex]1000\Omega[/tex]

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

[tex]q_{1} +q_{2} =-q_{3}[/tex] -----eqution 1

where,

[tex]q_{1}[/tex] is the heat absorbed by the solid at 0⁰C

[tex]q_{2}[/tex] is the heat absorbed by the liquid at 0⁰C

[tex]q_{3}[/tex] the heat lost by the warmer water sample

Important equations to be used in solving this problem

[tex]q=m *c*\delta {T}[/tex], where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

[tex]\delta {T}[/tex] is change in temperature

Again,

[tex]q=n*\delta {_f_u_s}[/tex] -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

[tex]=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O[/tex]

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

[tex]q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ[/tex]

This means that equation (1) becomes

79.13 KJ + [tex]q_{2} = -q_{3}[/tex]

Step 4: calculate the final temperature of the water

[tex]79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}[/tex]

Substitute in the values; we will have,

[tex]79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})[/tex]

79.13 kJ + 990.66J* [tex](T_{f}-218})[/tex] = -1463J*[tex](T_{f}-100})[/tex]

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* [tex](T_{f}-218})[/tex] = -1.463KJ*[tex](T_{f}-100})[/tex]

[tex]79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3[/tex]

collect like terms,

2.45366[tex]T_{f}[/tex] = 283.133

∴[tex]T_{f} =[/tex] = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

A solar heating system has a 25.0% conversion efficiency; the solar radiation incident on the panels is 1 000 W/m2.

What is the increase in temperature of 30.0 kg of water in a 1.00-h period by a 4.00-m2 -area collector? (cw = 4 186 J/kg⋅°C)

a. 14.3°Cb. 22.4°Cc. 28.7°Cd. 44.3°C

Answers

Answer:

c. 28.7C

Explanation:

Since the solar radiation incident on the panel is 1000W/m2 and the collector has an area of 4m2. We can conclude that the solar power generated is

1000 * 4 = 4000 W or J/s

Within 1 hour, or 3600 seconds this solar power generator should generate (with 25 % efficiency):

E = 0.25 * 4000 * 3600 = 3600000 J or 3.6 MJ

This energy will be converted to heat to heat up water. Using water heat specific of 4186 J/kg C we can find out how much temperature has raised:

[tex]\Delta T = \frac{E}{c*m} = \frac{3600000}{30 * 4186} = 28.7^oC[/tex]

So C. is the correct answer

A circuit consists of a coil that has a self-inductance equal to 4.3 mH and an internal resistance equal to 16 Ω, an ideal 9 V battery, and an open switch--all connected in series. At t = 0 the switch is closed. Find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil.

Answers

Answer:

t = 186.2 μs

Explanation:

Current in LR series circuit

[tex]I(t) = I_{s}( 1 - e^{-Rt/L)}[/tex]----(1)

steady current =  I_{s} = V/R

time constant = τ =[tex]L/R =4.3 * 10^{-3} / 16\\[/tex]

                                              = 0.268 ms

magnetic energy stored in coil = [tex]U_{L} = \frac{1}{2}LI^{2}[/tex]

rate at which magnetic energy stored in coil= [tex]\frac{d}{dt}U_{L} =\frac{d}{dt} \frac{1}{2}LI^{2}   \\                            = LI\frac{dI}{dt}\\[/tex]----(2)

rate at which power is dissipated in R:

                                       [tex]P = I^{2}R[/tex]---(3)

To find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil equate (2) and (3)

[tex]I^{2}R=LI \frac{dI}{dt}[/tex]

[/tex]I=\frac{L}{R}\frac{dI}{dt}[/tex]----(4)

differentiating (1) w.r.to t

[tex]I(t)=I_{f} (1-e^{\frac{Rt}{L} })[/tex]

[tex]\frac{dI}{dt} = I_{f}\frac{d}{dt}(1-e^{\frac{-Rt}{L} }   )[/tex]

[tex]\frac{dI}{dt}= I_{f}(-\frac{R}{L} e^{\frac{-Rt}{L} } )\\[/tex]---(5)

substituting (5) in (4)

[tex]I=I_{f}e^{-\frac{Rt}{L} }[/tex]----(6)

equating (1) and (6)

[tex]I_{f}( 1- e^{-\frac{Rt}{L} } ) = I_{f}e^{-\frac{Rt}{L} }[/tex]

[tex]1 - e^{-\frac{Rt}{L} } =  e^{-\frac{Rt}{L} }[/tex]

[tex]\frac{1}{2}= e^{-\frac{Rt}{L} }[/tex]

[tex]t= -\frac{L}{R}ln\frac{1}{2}[/tex]

L= 4.3 mH

R= 16 Ω

t = 186.2 μs

Suppose you are standing a few feet away from a bonfire on a cold fall evening. Your face begins to feel hot. What is the mechanism that transfers heat from the fire to your face? (Hint: Is the air between you and the fire hotter or cooler than your face?)

•A. convection
•B. radiation
•C. conduction
•D. none of the above

Answers

B. Radiation. It is not touching so it cannot be conduction

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.10 L ? (The temperature was held constant.)

Answers

Answer:

0.358g

Explanation:

Density of Helium = 0.179g/L

ρ=m/v

m=ρv

when the volume was 2L

m1= 0.179*2

m1=0.358g

when the volume increased to 4L

m2= 0.179*4

m2=0.716g

gram of helium added = 0.716g-0.358g

=0.358g

Two long straight wires enter a room through a window. One carries a current of 3.0 ???? into the room while the other carries a current of 5.0 ???? out. The magnitude in Tm of the path integral ∮ ????⃗ ∙ ???????? around the window frame is:

Answers

Answer:

[tex]\begin{equation}\\\oint_LB.dl\\\end{equation}[/tex] = -8πx[tex]10^{-7}[/tex]

Explanation:

If you need calculate

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex]

You can use the Ampere's Law

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = [tex]I_{in}[/tex]μ

   

      Where [tex]I_{in}[/tex]: Current passing through the window

                   μ : Free space’s magnetic permeability

                   μ = 4πx[tex]10^{-7} T.m.A^{-1}[/tex]

Then

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = (3-5)4πx[tex]10^{-7}[/tex]

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = -8πx[tex]10^{-7}[/tex]

The magnitude of in the path integral is 2.5*10^-6 T.M

Data;

I1 = 5A

I2 = 3A

μo = 4π * 10^-7 T.m.A^-1

Ampere Law

Ampere law states that the sum of the length of elements multiplied by the magnetic field in the path of the length elements is equal to the permeability multiplied by the electric current enclosed in the loop.

Mathematically;

∮B.dl = I*μo

I = current passing through the windowμo = free space magnetic permeability

[tex]\int B.\delta s = \mu I\\\int B.\delta s = \mu (5A - 3A)\\\int B.\delta s = 4\pi * 10^-^7 * 2A\\\int B.\delta s = 2.5*10^-^6 T.M[/tex]

The magnitude of in the path integral is 2.5*10^-6 T.M

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Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.

A. TrueB. False

Answers

Answer:

if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

2 eagles hit each other at a 90 degree angle they grab each other, what is their velocity and direction?

Answers

Answer:

This is an example of completely inelastic collision, since they grab each other and move as a single object after the collision.

The conservation of momentum requires

[tex]\vec{P_1} + \vec{P_2} = \vec{P}_{final}[/tex]

where

[tex]\vec{P_1} = m_1v_1 \^x\\\vec{P_2} = m_2v_2 \^y[/tex]

Their final momentum is

[tex]\vec{P}_{final} = m_1v_1 \^x + m_2 v_2 \^y[/tex]

Their final velocity and direction is

[tex]\vec{v}_{final} = \frac{m_1v_1}{m_1 + m_2}\^x + \frac{m_2v_2}{m_1 + m_2}\^y[/tex]

Although not states in the question, let's consider the case that the masses and speeds of the eagles are the same:

[tex]\vec{v}_{final} =  \frac{v}{2}\^x + \frac{v}{2}\^y[/tex]

A swimming pool heater has to be able to raise the temperature of the 40 000 gallons of water in the pool by 10.0 C°.

How many kilowatt-hours of energy are required?

(One gallon of water has a mass of approximately 3.8 kg and the specific heat of water is 4 186 J/kg⋅°C.)
a. 1 960 kWh
b. 1 770 kWh
c. 330 kWh
d. 216 kWh

Answers

Answer:

b. 1 770 kWh

Explanation:

The heat needed to change the temperature of a certain amount of a substance is given by:

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

Here m is the mass of the susbtance, C is the specific heat of the substance and [tex]\Delta T[/tex] is the temperature change

[tex]Q=(40000*3.8kg)(4186\frac{J}{kg\cdot ^\circ C})(10^\circ C)\\Q=6.36*10^9J[/tex]

Recall that one watt hour is equivalent to 1 watt (1 W) of power sustained for 1 hour. One watt is equal to 1 J/s. So, one watt hour is equal to 3600 J and one kilowatt hour is equal to [tex]3600*10^3 J[/tex]

[tex]Q=6.36*10^9J*\frac{1kW\cdot h}{3600*10^3J}\\Q=1766.66kW\cdot h[/tex]

Final answer:

To heat 40,000 gallons of water by 10.0 C° in a swimming pool, 1,767 kilowatt-hours of energy are required, rounding to the nearest so, option gives (b) 1,770 kWh as the answer.

Explanation:

The question asks: How many kilowatt-hours of energy are required to raise the temperature of 40,000 gallons of water in a pool by 10.0 C°? To solve this, we need to calculate the energy needed using the formula for heat energy: Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Firstly, convert the volume of water from gallons to kilograms. 40,000 gallons is approximately 40,000 x 3.8 kg = 152,000 kg. Next, use the specific heat of water (4,186 J/kg°C) and the temperature change (10.0 C°) to find the energy in joules: Q = 152,000 kg x 4,186 J/kg°C x 10.0 C° = 6,362,720,000 J.

To convert joules to kilowatt-hours, divide the total joules by 3,600,000 (the number of joules in one kilowatt-hour): 6,362,720,000 J / 3,600,000 J/kWh = 1,767 kWh. Therefore, the energy required is 1,767 kWh, making option (b) 1,770 kWh the nearest correct answer.

Which of the following statements correctly describes the law of conservation of energy? Group of answer choicesa. Mass cannot be created but it can be destroyed under extreme pressures.b. Mass cannot be conserved during a chemical reaction; a little bit of mass is always lost.c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed.d. When added to a system, energy can destroy mass.

Answers

To solve this problem we will also apply the concept related to the conservation of the mass, which announces that: "In an isolated system, during any ordinary chemical reaction, the total mass in the system remains constant, that is, the mass consumed by the reagents is equal to the mass of the products obtained. "

If the mass is in a closed system, it cannot change. This assessment should not be confused with the transformation of the matter within it, for which it is possible that over time the matter will change from one form to another. For example during a chemical reaction, there is a rupture of links to reorganize into another, but said mass in the closed system is maintained.

The correct answer is:

C. "The mass of a closed system cannot change over time; mass cannot be created or destroyed."

The following statements correctly describe the law of conservation of energy - c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed

The law of conservation of mass states that the mass is an isolated system that can not be created nor destroyed.

conserved means saved, so according to the law of conservation of mass refers to the "saving" of mass.

Thus, The following statements correctly describe the law of conservation of energy - c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed

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