While Dr. Chesnutt is making breakfast, she turns on her 1060 W toaster, 500 W coffee pot, and 1230 W microwave at the same time. Part Description Answer Save Status A. If all three appliances are connected in parallel across a 120 V power source, what total current would they draw from the source? (include units with answer) Format Check Click here to check your answer 4 pts.100% 13% try penalty 8 try limit # tries: 0

Answers

Answer 1

Answer:23.25 A

Explanation:

Given

Rating of toaster [tex]P_1=1060\ W[/tex]

Coffee pot [tex]P_2=500\ W[/tex]

microwave [tex]P_3=1230\ W[/tex]

Voltage applied [tex]V=120\ V[/tex]

if they are connected in parallel then all three operates at same voltage

so their resistance are

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

thus [tex]R_1=\frac{V^2}{P_1}=\frac{120^2}{1060}[/tex]

[tex]R_1=13.58\ \Omega[/tex]

[tex]R_2=\frac{V^2}{P_2}=\frac{120^2}{500}[/tex]

[tex]R_2=28.8\ \Omega[/tex]

[tex]R_3=\frac{V^2}{P_3}=\frac{120^2}{1230}[/tex]

[tex]R_3=11.707\ \Omega[/tex]

and [tex]V=IR[/tex]

where I=current

thus [tex]I_1=\frac{V}{R_1}=\frac{120}{13.58}[/tex]

[tex]I_1=8.83\ A[/tex]

[tex]I_2=\frac{V}{R_2}=\frac{120}{28.8}[/tex]

[tex]I_2=4.16\ A[/tex]

[tex]I_3=\frac{V}{R_3}=\frac{120}{11.707}[/tex]

[tex]I_3=10.25\ A[/tex]

Total current [tex]I=I_1+I_2+I_3=23.24\ A[/tex]


Related Questions

The hydronium ion concentration in an aqueous solution is [H+] = 5.3 x10-4 M. What is the hydroxide ion concentration? Is this solution acidic or basic?

A.) [OH-] = 1.9 x10-11 M, acidic

B.) [OH-] = 4.7 x10-6 M, basic

C.) [OH-] = 1.1 x10-9 M, acidic

D.) [OH-] = 3.2 x10-8 M, acidic

Answers

Answer:a

Explanation:

Given

[tex][H^+]=5.3\times 10^{-4}\ M[/tex]

and [tex]pH+pOH=14[/tex]

Also [tex]pH=-\log [H^+][/tex]

therefore [tex]pH=-\log (5.3\times 10^{-3})[/tex]

[tex]pH=-(-4-\log (5.3))[/tex]

[tex]pH=3.275[/tex]

Thus [tex]pOH=14-3.275[/tex]

[tex]pOH=10.275[/tex]

and [tex]pOH=-\log [OH^{-}][/tex]

[tex][OH]^{-1}=10^{-10.25}[/tex]

[tex][OH]^{-1}=1.88\times 10^{-11}\ M[/tex]

As the pH is less than 7 therefore solution is acidic

A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground. Given that her heavier child of weight WWW is sitting a distance LLL to the left of the pivot, at what distance L1L1L_1 must she place her second child of weight www on the right side of the pivot to balance the seesaw? Express your answer in terms of LLL, WWW, and www.

Answers

Answer:

L1L1L_1 = ( LLL × WWW)/ www.

Explanation:

From the question we are given that the children's weight are not equal, thus, there will be a need to balance the seesaw "so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground''.

Also, From the question, we are gven that the weight of the heavier child = WWW, the heavier child's is sitting a distance = LLL, the weight of the second child= www.

Therefore, in order to balance the seesaw, she will need to place her second child of weight www on the right side of the pivot to balance the seesaw at a distance of;

L1L1L_1 = ( LLL × WWW)/ www.

To understand the formula representing a traveling electromagnetic wave. Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagnetic waves. Electromagnetic waves comprise combinations of electric and magnetic fields that are mutually compatible in the sense that the changes in one generate the other. The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given by

Answers

Answer:

E(x,t) = Emaxcos(kx - ωt + φ),

B(x,t) = Bmaxcos(kx - ωt + φ).

Explanation:

E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves the electric field E and the magnetic field B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E x B.

Compared to the current in the 10.-ohm resistance in the circuit shown at the right, the current in the 5.0-ohm resistance is
a. one-half as great
b. one-fourth as great
c. the same
d. twice as great

Answers

Answer:

The answer is d. Twice as great.

Explanation:

The German physicist and mathematician Georg Simon Ohm says in his basic law of electrical circuits or Ohm's law that the potential difference V that is applied at the ends of a conductor is proportional to the intensity I of the current that circulates and that the electrical resistance R is the ratio factor between I and V.

The equation would be I = V / R

If by any chance R is reduced because it is inversely proportional, the current would increase.

So if the current goes from 10 to half 5, its current would double.

If the temperature of this balloon were to decrease suddenly, how would the balloon change?

A. Its mass would increase.

B. Its mass would decrease.

C. Its volume would increase.

D. Its volume would decrease.

Answers

D. it’s volume would decrease

Final answer:

The volume of a balloon decreases when its temperature is suddenly lowered, in accordance with Charles's Law, because the gas particles inside the balloon move less vigorously due to a decrease in their average kinetic energy.

Explanation:

When the temperature of a balloon decreases suddenly, according to Charles's Law, the volume of the gas inside the balloon also decreases, assuming that all other conditions, such as pressure and the amount of gas, remain constant. This occurs because a decrease in temperature leads to a decrease in the average kinetic energy of the gas particles, meaning they move less vigorously and occupy less space. Therefore, the correct answer to the question would be D. Its volume would decrease.

Calculate the induced electric field (in V/m) in a 52-turn coil with a diameter of 12 cm that is placed in a spatially uniform magnetic field of magnitude 0.30 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)

Answers

Given Information:  

Number of turns = N = 52

Diameter of coil = d = 12 cm = 0.12 m

Time = t = 0.10 seconds

Magnetic field = B = 0.30 T  

Required Information:  

Induced electric field = E = ?  

Answer:

Induced electric field = E = 4.68 V/m

Explanation:

The Maxwell's third equation can be used to find out the induced electric field,

∫E.dl = -dΦ/dt  

Where E is the induced electric field, dl is the circumference of the loop and dΦ/dt  is the rate of change of magnetic flux and is given by

Φ = NABcos(θ)

Where N is the number of turns, A is the area of coil and B is the magnetic field and cos(θ) = 1

Φ = NAB

∫E.dl = -dΦ/dt  

E(2πr) = -d(NAB)/dt

E =1/(2πr)*-d(NAB)/dt

E =NA/(2πr)*-dB/dt

Area is given by

A = πr²

E =Nπr²/(2πr)*-dB/dt

E =Nr/2*-dB/dt

The magnetic field reduce from 0.30 to zero in 0.10 seconds

E =Nr/2*-(0.30 - 0)/(0 - 0.10)

E =Nr/2*-(0.30)/(-0.10)

E = Nr/2*-(-3)

The radius r is given by

r = d/2 = 0.12/2 = 0.06 m

E = (52*0.06)/2*(3)

E = 1.56*3

E = 1.56*3

E = 4.68 V/m

Therefore, the induced electric field in the coil is 4.68 V/m

g A simple pendulum (consisting of a point mass suspended by a massless string) on the surface of the earth has a period of 1.00 s. On a distant planet, a pendulum of identical length but twice the mass is found to have the same period. What is true about the acceleration due to gravity on the distant planet? [Assume both pendulums operate free of any friction or air resistance force]

Answers

Answer:

Explanation:

The formula for time period of a pendulum is given as follows :

T = 2π[tex]\sqrt{\frac{l}{g} }[/tex]

l is length of pendulum and g is acceleration due to gravity .

So time period of pendulum  is not dependent on the mass of the pendulum . If time period is same and length is also the same then acceleration due to gravity will also be the same . Hence the acceleration due to gravity at distant planet will be same as that on the earth.

A hydrogen atom can be in the 1S state, whose energy we'll call 0, the 2S state, or any of 3 2P states. The 2S and 2P states have energies of 10.2 eV. There are other states with higher energy but we'll ignore them for simplicity. The 2P states have distinctive optical properties, so we're interested in how many are present even when it's a small fraction of the total. 1) What fraction of the H is in 2P states at T=5900 K, a typical Sun surface temperature?

Answers

Final answer:

The fraction of hydrogen atoms in the 2P states at T=5900 K is calculated using the Boltzmann distribution. The energy of the 2S and 2P states is 10.2 eV, and their degeneracies are factored into the Boltzmann factor to determine the relative populations of these states.

Explanation:

The fraction of hydrogen atoms in the 2P states at a temperature of 5900 K can be calculated using the Boltzmann distribution. According to quantum mechanics, the energy levels of a hydrogen atom involve principal quantum numbers, with the ground state being 1s and higher energy excited states being designated by higher quantum numbers and corresponding letters (s, p, d, f, etc.) for their angular momentum quantum numbers. Since we are given that the energy of 2S and 2P states are the same at 10.2 eV and the ground state (1S) has an energy we'll call 0, we can use the Boltzmann factor to find the relative populations of these states.

To calculate the fraction of hydrogen atoms in the 2P state, use the following Boltzmann factor equation: fraction = (g2P / gTotal) * exp(-E2/kT), where g2P is the degeneracy of the 2P state, gTotal is the total degeneracy of all states considered, E2 is the energy of the 2P state, k is the Boltzmann constant, and T is the temperature. The degeneracy of the 2P state is 3 (since there are three 2P states) and the only other state considered here is the 2S state, which has degeneracy 1, making gTotal = 4. Plugging in the energy of 10.2 eV for E2 and converting it to joules (multiply by 1.602 x 10-19 J/eV), using k = 1.38 x 10-23 J/K, and T = 5900 K, we can calculate the fraction of hydrogen in the 2P state.

A spring hangs from the ceiling with an unstretched length of x0=0.45 m . A m1=7.9 kg block is hung from the spring, causing the spring to stretch to a length x1=0.74 m.Find the length x2of the spring when a m2=3.3 kg block is hung from the spring. For both cases, all vibrations of the spring are allowed to settle down before any measurements are made.

Answers

Final answer:

To find the new length of the spring when a different mass is hung from it, calculate the spring constant using the initial mass, and then use Hooke's Law to solve for the extension caused by the new mass. Add this extension to the original unstretched length of the spring to find the stretched length.

Explanation:

The student's question involves finding the new length of a spring x2 when a block with mass m2 of 3.3 kg is hung from it. Given an initial unstretched length (x0) and a stretched length (x1) with block m1, we can solve for the spring constant k using Hooke's Law, which states F = k * x, where F is the force, k is the spring constant, and x is the extension from the natural length of the spring.

First, we calculate the extension caused by m1: extension_m1 = x1 - x0 = 0.74 m - 0.45 m = 0.29 m. The force exerted by m1 is F1 = m1 * g, where g is the acceleration due to gravity (9.81 m/s2). Calculating the force, we get F1 = 7.9 kg * 9.81 m/s2. Then, the spring constant k can be found using k = F1 / extension_m1.

With the value of k, we can find the extension caused by m2: extension_m2 = F2 / k, where F2 = m2 * g. Finally, the length x2 of the spring with m2 hung from it is x2 = x0 + extension_m2.

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?

Answers

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law,  PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

[tex]V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}[/tex]

Given;

[tex]\frac{dP}{dt}[/tex] (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) =  600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( [tex]\frac{dv}{dt}[/tex]);

(600 x 40) + (150 x [tex]\frac{dv}{dt}[/tex]) = 0

[tex]\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min[/tex]

Therefore, the volume is decreasing at 160 cm³/min

Final answer:

Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C. To find the rate at which the volume is decreasing at a given instant, we differentiate the equation PV = C with respect to time and substitute the given values.

Explanation:

Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C, where P is the pressure, V is the volume, and C is a constant. In this scenario, the pressure is increasing at a rate of 40 kPa/min. To find the rate at which the volume is decreasing at this instant, we differentiate the equation PV = C with respect to time. Applying the chain rule, we get:

d(PV)/dt = d(C)/dt

P(dV/dt) + V(dP/dt) = 0

dV/dt = - (V/P)(dP/dt)

We substitute the given values where the volume V = 600 cm3 and pressure P = 150 kPa:

dV/dt = - (600 cm3 / 150 kPa) (40 kPa/min)

dV/dt = -160 cm3/min

Therefore, the volume is decreasing at a rate of 160 cm3/min at this instant.

the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet

Answers

Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

Answer:

The force that is exerted when a shopping cart is pushed:

-Contact

The force that causes a metal ball to move toward a magnet:

-Noncontact

The expression known as​ Froude's number is given as Upper F equals StartFraction v squared Over gl EndFraction . This number was used in shipbuilding and also in research on dinosaurs. In this​ example, l is given as well as​ Froude's number. Find the​ velocity, v​ (in meters per​ second) of a rhinoceros. It is known that gequals9.8 meters per second squared. ​Rhinoceros: lequals​ 1.2; Froude number equals 2.6

Answers

Answer:[tex]v=5.529\ m/s[/tex]

Explanation:

Given

number is given as [tex]=\frac{v^2}{gl}[/tex]

and [tex]number=2.6[/tex]

[tex]l=1.2\ m[/tex]

g=acceleration due to gravity[tex](9.8\ m/s^2)[/tex]

calculating for [tex]v[/tex]

[tex]v^2=gl(2.6)[/tex]

[tex]v=\sqrt{2.6\times 9.8\times 1.2}[/tex]

[tex]v=\sqrt{30.576}[/tex]

[tex]v=5.529\ m/s[/tex]

A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively. (a) Find the energy stored in the inductor when the current reaches its maximum value. J (b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch

Answers

Answer:

a) 20.81 J

b) 8.29 J

Explanation:

V = iR + L di/dt

where

i = a(1-e^-kt)

for large t

i = V/R

i = 24 / 9.4

i = 2.55 A

so

i = 2.55(1-e^-kt)

di/dt = 2.55 k e^-kt

24 = 24-24e^-kt + 6.4(2.55)k e^-kt

24 = 6.4(2.55) k

k = 24 / (6.4 * 2.55)

k = 24 / 16.32

k = 1.47 = R/L

so

i = 2.55(1-e^-(Rt/L))

current is maximum at great t

i max = 2.55 - 0

energy = (1/2) L i^2

E = (1/2)(6.4)2.55^2

E = 20.81 Joules

one time constant T = L/R and e^-(Rt/L) = 1/e = .368

i = 2.55 (1 - 0.368)

i = 2.55 * 0.632

i = 1.61 amps

energy = (1/2)(6.4)1.61^2

E = 8.29 Joules

The energy stored in the inductor when the current reaches its maximum value is 20.80 J. After one time constant, the energy stored is approximately 8.25 J.

we need to determine the energy stored in the inductor at two different points in time.

a) The energy stored in an inductor is given by the formula:

U = (1/2) × L × I²

Maximum Current

In an RL circuit, the maximum current is achieved when the circuit reaches a steady state. The maximum current is given by:

I_max = V/R, where V is the voltage and R is the resistance.

Given V = 24.0 V and R = 9.40 Ω, we have:

I_max = 24.0 V / 9.40 Ω = 2.55 A

Now, the energy stored in the inductor at the maximum current is:

U_max = (1/2) × L × I_max²

Given L = 6.40 H, we get:

U_max = (1/2) × 6.40 H × (2.55 A)² = 20.80 J

b)   Energy After One Time Constant

The time constant for an RL circuit is defined as:

τ = L/R

Given L = 6.40 H and R = 9.40 Ω, we calculate:

τ = 6.40 H / 9.40 Ω ≈ 0.68 s

At one time constant, the current has reached approximately 63.2% of its maximum value:

I_τ ≈ 0.632 × I_max = 0.632 × 2.55 A ≈ 1.61 A

The energy stored in the inductor at this time is:

U_τ = (1/2) × L × I_τ² = (1/2) × 6.40 H × (1.61 A)² ≈ 8.25 J

complete question:

A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively.

(a) Find the energy stored in the inductor when the current reaches its maximum value. J

(b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch  is closed. j

In aircraft design, the pressure coefficient Cp is usually measured during wind tunnel testing of an aircraft component to predict structural loads, as well as lift and drag forces. Let Cp=(p−p[infinity])/ (0.5*rhoU[infinity]^2)where U[infinity] is the airspeed of the test at pressure p[infinity] and using a fluid of density rho. The pressure difference p ‐ p[infinity] is measured directly where p is the pressure located at a point on the surface of the test component. Estimate the uncertainty in Cp under the following conditions: Δp = p − p[infinity] = 1,000 N/m2, uΔp = 15 N/m2; rho = 1.20 kg/m3, urho = 0.01 kg/m3; U[infinity] = 50.0 m/s, u_U[infinity] = 0.21 m/s. Assume 95% confidence in stated values.

Answers

Answer:

Check the explanation

Explanation:

From given data, it can be noted that 95% of given confidently data, means 5% of data is uncertain. According to the question, we have to calculate uncertainty in Cp .

 Kindly check the attached image below for the step by step explanation to the question above.

how much work is done to move 3c of charge through a potential difference of 1.5v

Answers

Answer:4.5 joules

Explanation:

charge=3c

Potential difference=1.5v

Work=potential difference x charge

Work=1.5 x 3

Work=4.5

Work=4.5 joules

To understand how to use the principle of superposition in conjunction with the Biot-Savart (or Ampere's) law. From the Biot-Savart law, it can be calculated that the magnitude of the magnetic field due to a long straight wire is given by Bwire=μ0I2πd , where μ0 (=4π×10−7T⋅m/A) is the permeability constant, I is the current in the wire, and d is the distance from the wire to the location at which the magnitude of the magnetic field is being calculated. The same result can be obtained from Ampere's law as well.

Answers

Final answer:

The principle of superposition can be used in conjunction with the Biot-Savart (or Ampere's) law to calculate the magnetic field due to a combination of current-carrying wires. Ampere's law is a more general law that relates the magnetic field to the total current passing through a closed loop.

Explanation:

The principle of superposition can be used in conjunction with the Biot-Savart (or Ampere's) law to calculate the magnetic field due to a combination of current-carrying wires. The Biot-Savart law allows us to calculate the magnetic field at any point due to an element of current in a wire. By integrating this law over the entire length of the wires and applying the principle of superposition, we can determine the total magnetic field produced by multiple wires.

For example, if we have two parallel wires carrying currents I1 and I2, the total magnetic field at a point due to both wires can be found by summing the individual magnetic fields produced by each wire using the Biot-Savart law.

It is important to note that Ampere's law can also be used to determine the magnetic field produced by current-carrying wires. It is a more general law that relates the magnetic field to the total current passing through a closed loop. The results obtained from Ampere's law are consistent with the Biot-Savart law.

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Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off?

Answers

Final answer:

The angular speed of the merry-go-round would increase when the boy jumps off due to the conservation of angular momentum. However, without specific values like the weights and distances involved, the precise speed cannot be calculated.

Explanation:

The speed of the merry-go-round after the boy jumps off cannot be determined from the given information. However, we can use the principle of conservation of angular momentum to analyze the situation; this principle states that the initial angular momentum (while the boy was still on the merry-go-round) should be equal to the final angular momentum (after the boy jumps off). The angular momentum is the product of moment of inertia and angular speed. When the boy was sitting on the edge of the merry-go-round, he had a certain moment of inertia due to his mass and the distance from the center. Once he jumps off, the moment of inertia decreases, hence according to the conservation law, the angular speed must increase to keep the momentum constant.

However, the exact increase in angular speed could only be calculated if we knew the weight of the boy and of the merry-go-round, as well as the radius of the merry-go-round (to calculate moments of inertia), and the initial angular speed. Another simplification made in this situation is that the friction between the merry-go-round and its axle is negligible, which might not be true in real life scenarios.

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Final answer:

When the boy jumps off the merry-go-round, the total angular momentum of the system should conserve. As the boy's leaving reduces the moment of inertia, the angular speed of the merry-go-round must increase to preserve angular momentum. To calculate the exact increase, we would need specific values for the moment of inertia before and after the jump.

Explanation:

The question asked is associated with angular velocity and moment of inertia in the context of a merry-go-round. As the boy jumps off the merry-go-round, the system's total angular momentum must be conserved, because there are no external torques acting on it.

When the boy was still on the merry-go-round, he was part of the system, contributing to the total angular momentum. But when he departs, he no longer contributes to the system's moment of inertia, effectively reducing it. As a result, the angular speed of the merry-go-round must increase to conserve angular momentum (since angular momentum = moment of inertia x angular speed).

Unfortunately, without specific values for the moment of inertia both before and after the boy jumps off, we can't calculate the exact increase in angular speed. However, the key concept is the conservation of angular momentum: when the moment of inertia decreases, the angular speed must increase to maintain constant total angular momentum.

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In deserts the temperatures can reach extreme values in deep of summer and winter. A steel bridge is being constructed and it is supposed to be 1234.567 m long at 233.15 K in the coldest possible winter. What will be the possible length of this bridge when the temperature rises to a possible extreme of +140.0°F in hottest part of summer? The average coefficient of linear expansion of the steel is 11.0123 × 10-6 K-1.

Answers

Answer:

The length of the bridge during the hottest part of summer is [tex]L_s = 1235.925 m[/tex]

Explanation:

From the question we are told that

    The length of the steel bridge is [tex]L = 1234.567m[/tex]

     The temperature for this length is [tex]T_1 = 233.15K[/tex]

     The temperature at summer [tex]T_2 = + 140.0F = \frac{140 - 32}{180} *100 + 273= 333 K[/tex]

     The coefficient of linear expansion is [tex]\alpha = 11.0123*10^{-6} K^{-1}[/tex]

Generally the change in length of the steel bridge is mathematically represented as

             [tex]\Delta L = \alpha L \Delta T[/tex]

Substituting value

             [tex]\Delta L = 11.0123*10^{-6} * 1234.567 (333-233.15)[/tex]

              [tex]\Delta L = 1.3575 \ m[/tex]

The length of the bridge in summer is mathematically evaluated as

         [tex]L_s = L + \Delta L[/tex]

Substituting values

         [tex]L_s = 1234.567 + 1.3575[/tex]

        [tex]L_s = 1235.925 m[/tex]

In a shipping company distribution center, an open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. You can ignore friction between the cart and the floor. A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal. The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds. The package comes to a stop in the cart after 4 seconds. What is:

Answers

Answer:

(1). 9.35 m/s.

(2).V2 = 3.22 m/s to the left.

Explanation:

So, we are given the following data or parameters or infomation in the question above as;

(1). "open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. "

(2). "A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal."

(3). "The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds."

(4). "The package comes to a stop in the cart after 4 seconds."

(a). So, in order to solve this question, we will be making use of the equations below;

(1). Gravitational potential energy = mass × acceleration due to gravity × height.

(2). Kinetic energy = 1/2 × mass × (velocity)^2.

(3). Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).

So, starting from equation (1) above;

Gravitational potential energy = mass × acceleration due to gravity × height.

Gravitational potential energy = 15 × 9.8 × 4 = 588 J.

Kinetic energy = 1/2 × mass × (velocity)^2. = 1/2 × 15 × (3)^2.= 67.5 J.

kinetic energy (final) = 1/2 × 15 × v^2. = 7.5v^2.

Hence, Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).

[potential energy (final) = 0].

67.5 + 588 = 7.5v^2.

v = 9.35 m/s

(b). -3 cos (27°) = -2.67 m/s.

15 × -2.67 + 50 × 5 =( 15 + 50) × v2.

=> 209.5 = 65v2.

=> V2 = 3.22 m/s to the left.

(1) The speed of the package before it lands on the cart is 9.35 m/s.

(2) The final speed of the cart is 3.22 m/s

Conservation of energy and momentum:

The gravitational potential energy of a system is given by

PE = mgh

where m is the mass

g is the acceleration due to gravity

and h is the height.

The kinetic energy is given by:

KE = (1/2) mv²

According to the law of conservation of energy:

Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).

The final potential energy will be zero since taking the cart as the reference as ground.

PE(initial) = 15 × 9.8 × 4 = 588 J

KE(initial) = (1/2) mv² = 1/2 × 15 × 3² = 67.5 J

KE(final) = 1/2 × 15 × v² = 7.5v²

So,

67.5 + 588 = 7.5v²

v = 9.35 m/s

(b) The package leave at an angle of 27° down the horizontal with velocity 3 m/s, so the horizontal speed is:

v ' = -3 cos (27°) = -2.67 m/s.

Negative sign, as the package is moving opposite to the cart

After the package lands on the cart, both move together, so from conservation of momentum:

15 × (-2.67) + 50 × 5 =( 15 + 50) × V

where V is the speed of cart and package together

209.5 = 65 V

V = 3.22 m/s to the left.

The question was incomplete. After searching it, it is likely that it was as given below:

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 27∘ from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what is (1) the speed of the package just before it lands in the cart and (2) the final speed of the cart?

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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 75.0-g piece of copper at 60.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20.0°C.

Determine the specific heat of the unknown sample.

Answers

Answer:

The specific heat of the unknown sample is 1822.14 J/kg.k

Explanation:

Given;

mass of aluminum calorimeter, [tex]M_c[/tex] = 100 g

mass of water, [tex]M_w[/tex] = 250 g

stabilizing temperature of water-calorimeter, ΔT =  10.0°C

mass of copper, [tex]M_c_u[/tex] = 75 g

initial temperature of copper, [tex]T_{cu}[/tex] =  60.0°C

mass of unknown sample, [tex]M_u[/tex] = 70.0 g

initial temperature of unknown sample, [tex]T_u[/tex] =  100°C.

The final temperature of the entire system, t = 20.0°C

Apply the principle of conservation of energy;

energy used to heat water and calorimeter is equal to energy released by copper and unknown sample.

[tex]Q = M_{cu}C_{cu} \delta T_{cu} + M__{u}C{_u} \delta T_u[/tex]

where;

Q is energy used to heat water and calorimeter

[tex]C_c_u[/tex] is the specific heat capacity of copper

[tex]C_u[/tex] is the specific heat capacity of unknown sample

Make [tex]C_u[/tex] subject of the formula;

[tex]C_u = \frac{Q-M_c_u C_c_u \delta T_c_u}{M_u \delta T_u} \\\\C_u = \frac{(C_wM_w +C_cM_c)\delta T-M_c_u C_c_u \delta T_c_u}{M_u \delta T _u} \\\\C_u = \frac{(4186*0.25 +900*0.1)10-0.075* 387 *40}{0.07* 80} \\\\C_u = \frac{11365 -1161}{5.6} \\\\C_u = 1822.14 \ J/kg.k[/tex]

Therefore, the specific heat of the unknown sample is 1822.14 J/kg.k

The specific heat of the unknown sample is;

c_u = 1823 J/Kg.k

We are given;

Mass of Aluminum calorimeter; m_c = 100 g = 0.1 kg

Mass of water; m_w = 250 g = 0.25 kg

Initial temperature of Calorimeter and water; T_c = T_w = 10°C = 283 K

Mass of Copper; m_cu = 75 g = 0.075 kg

Initial temperature of Copper; T_cu = 60°C = 333 K

Mass of unknown sample; m_u = 70 g = 0.07 kg

Initial temperature of unknown substance = 100°C = 373 K

Final temperature of system; T_f = 20°C = 293 K

Formula for quantity of heat is;

Q = mcΔt

where;

m is mass

c is specific heat capacity

Δt is change in temperature;

For the calorimeter and water , we have;

Q_cw = (m_w*c_w + m_c*c_c)Δt

Specific heat capacity of water is; c_w = 4186 J/Kg.K

Specific heat capacity of aluminium is 900 J/Kg.K

Thus;

Q_cw = ((0.25 * 4186) + (0.1 * 900))(293 - 283)

Q_cw = 11365 J

For the unknown sample and the piece of copper;

Q_cu,u = (m_cu*c_cu*Δt) + (m_u*c_u*Δt)

specific heat capacity of copper; c_cu = 385 J/Kg.K

Thus;

Q_cu,u = (0.075*385*(333 - 293)) + (0.07*c_u*(373 - 293))

Q_cu,u = 1155 + 5.6c_u

From conservation of energy principle;

Q_cw = Q_cu,u

Thus;

11365 = 1155 + 4.2c_u

c_u = (11365 - 1155)/5.6

c_u = 1823 J/Kg.k

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This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force wil At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If q is positive, what is the direction of the force on the particle due to the magnetic field?

Answers

Answer:

force is  -y direction

Explanation:

The magnetic force is given by the equation

        F = q v x B

where bold letters indicate vectors.

The direction of this force can be found by the right-hand rule, where the thumb points in the direction of speed, the other fingers extend in the direction of the magnetic field and the palm is in the direction of force for a positive charge

let's apply this to our case

the speed is in the direction + x the thumb

the magnetic field in the + z direction fingers extended

the palm is in the direction - y

force is  -y direction

Final answer:

The force due to the magnetic field on a positively charged particle moving in the +x direction and perpendicular to the magnetic field will be in the +-y direction.

Explanation:

At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If q is positive, the direction of the force on the particle due to the magnetic field will be in the +-y direction.

A bungee jumper jumps from a tall bridge that is 60m above a flat, dry creek bed. She is carrying an audio generator that emits sound at frequency 1200 Hz. The unstretched length of the bungee cord is 26m, and the spring constant is selected so that she stops just before hitting the creek bed. At what point in her fall is the beat frequency, determined by listening to the audio generator and the reflection from the creek bed, a maximum

Answers

Answer:

Explanation:

The original frequency of sound  f₀

The apparent frequency of sound fa

For apparent frequency the formula is

fa =  [tex]f_0\times\frac{V+v}{V-v }[/tex]     , v is velocity of jumper which increases as he goes down .

Beat frequency

= fa - f₀

=  [tex]f_0\times(\frac{V+v}{V-v }-1)[/tex]

=  [tex]f_0\times(\frac{2v}{V-v })[/tex]

since v is very small in comparison to V , velocity of sound , in the denominator , v can be neglected.

beat frequency = [tex]f_0\times(\frac{2v}{V })[/tex]

v , the velocity of jumper will go on increasing as long as net force on the jumper is positive or

mg > kx where x is extension in the cord and k is its force constant . Below this point kx or restoring force becomes more than weight of the jumper and then net force on the jumper directs upwards. At this point beat frequency becomes maximum.

What measurement stays the same at any point in a parallel circuit?

A. current
B. voltage
C. resistivity
D. power

Answers

Answer:Voltage

Explanation:

Voltage stays the same at any point in a parallel circuit

A plane circular loop of conducting wire of radius with possesses turn is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength Bis increased at a constant rate from to in a time interval of 10s. (a) What is the emf generated around the loop? (b) If the electrical resistance of the loop is , what current flows around the loop as the magnetic field is increased?

Answers

Complete question;

A plane circular loop of conducting wire of radius r-10 cm which possesses 15 turns is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength is increased at a constant rate from IT to 5T in a time interval of 10 s.

a) What is the emf generated around the loop?

b) If the electrical resistance of the loop is 15Ω, what current flows around the loop as the magnetic field is increased?

Answer:

A) E.M.F generated around loop = 0.163 V

B)Current in loop; I = 0.011 A

Explanation:

A) We are given;

Initial magnetic field strength;B1 = 1T

Final magnetic field strength;B2 = 5T

Number of turns;N = 15 turns

Radius; r = 10cm = 0.1m

Angle;θ = 30°

Time interval; Δt = 10 s

Now, the formula for magnetic flux is: Φ = NABcosθ

Where;

N is number of turns

A is area = πr²

B is magnetic field strength

θ is angle

So, initial magnetic flux is;

Φ1 = NA(B1)cosθ

Plugging in the relevant values to obtain;

Φ1 = 15*(π*0.1²)(1)cos30

Φ1 = 0.4081 Wb

Similarly, final magnetic flux is;

Φ2 = NA(B2)cosθ

Plugging in the relevant values to obtain;

Φ2 = 15*(π*0.1²)(5)cos30

Φ2 = 2.0405 Wb

The time rate of change of the flux is;

dΦ_B/dt = (Φ2 - Φ1)/Δt

So, dΦ_B/dt = (2.0405 - 0.4081)/10

dΦ_B/dt = 0.163 Wb/s

Thus, the emf generated around the loop is; E = dΦ_B/dt = 0.163 V

B) from Ohm's law, the current which flows around the loop in response to the emf is given as;

I = E/R

We are given R =15Ω

Thus; I = 0.1632/15

I = 0.011 A

The three small spheres are welded to the light rigid frame which is rotating in a horizontal plane about a vertical axis through O with an angular velocity w= 20 rad/s. If a couple MO= 30 N.m is applied to the frame for 5 seconds, compute the new angular velocity.

Answers

Final answer:

The new angular velocity is computed using the kinematic expression w² = wo² + 2a0, where 'wo' is the original angular velocity 'a' is the angular acceleration and 'w' is the new angular velocity. The calculation is guided by principles of physics primarily involving angular momentum and acceleration.

Explanation:

The calculation for the new angular velocity can be gleaned from the kinematic expression w² = wo² + 2a0. The formula described represents the angular motion of the frame. The angular velocity originally is 20 rad/s and a couple MO of 30 N.m is applied to the frame for 5 seconds. The development of this concept involves principals of angular momentum as well as angular acceleration.

The angular acceleration can be calculated using the relation a = nett

Bearing in mind the initial angular velocity, the applied couple MO, and the duration of its application, the new angular velocity can be computed using the given formula. Note that the laws of physics, specifically the law of conservation of angular momentum, play an essential part in this calculation.

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A 0.060-kg ice hockey puck comes toward a player with a high speed. The player hits it directly back softly with an average force of 1.50 x 10^3 N. The hockey stick is in contact with the ball for 1.20 ms, and the ball leaves the stick with a velocity of 8.00 m/s. Let the direction of the force be the + x direction. Find the following (note: be careful with the sign/direction of the values):_______. 1. The final momentum of the ball 2. The impulse on the ball 3. The initial velocity of the ball

Answers

Answer:u=-22 m/s

Explanation:

Given

mass of puck [tex]m=0.06\ kg[/tex]

Average force [tex]f_{avg}=1.5\times 10^3\ N[/tex]

time of contact [tex]t=1.2ms=1.2\times 10^{-3}\ s[/tex]

puck leaves with a velocity of [tex]v=8\ m/s[/tex]

We know impulse is [tex]F_{avg}\Delta t[/tex][tex]=\text{change in momentum}[/tex]

therefore

[tex]1.5\times 10^3\times (1.2\times 10^{-3})=P_f-P_i[/tex]

[tex]P_i=0.06\times 8-1.8[/tex]

[tex]P_i=0.48-1.8=-1.32\ kg-m/s[/tex]

Final momentum [tex]P_f=m\times v_f[/tex]

[tex]P_f=0.06\times 8[/tex]

[tex]P_f=0.48\ kg-m/s[/tex]

Impulse on the ball [tex]=F_{avg}\Delta t[/tex]

Impulse[tex]=1.5\times 10^3\times 1.2\times 10^{-3}=1.8\ N-s[/tex]

Initial velocity is given by

[tex]u=\frac{P_i}{m}=\frac{-1.32}{0.06}[/tex]

[tex]u=-22\ m/s[/tex]

i.e. initially ball is moving towards -x-axis

A very long, straight solenoid with a cross-sectional area of 2.16 cm^2 is wound with 85.7 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)= (( 0.175 A/s^2 )t^2). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?

Answers

Answer:

1.74x10⁻⁵ V

Explanation:

n = 85.7 turns/cm => 8570 turns/metre

The field inside the long solenoid is given by B = μ₀ni

B =  4πx10⁻⁷ x 8570 x 0.175t² = 1.884x10⁻³ t²

dB/dt = 3.78x10⁻³ t

 

Cross-sectional Area'A'= 2.16 cm²=> 2.16 x [tex]10^{-4}[/tex] m²

Now, rate of change of flux linkage  '|Emf|' is given by:

|Emf| = d(NAB)/dt = NA dB/dt

|Emf|  = 5 x 2.16 x [tex]10^{-4}[/tex] x 3.78x10⁻³ t

|Emf| = 4.0824x10⁻⁶ t

Considering time 't' at which the current = 3.2A , we have

3.2 = 0.175T²

T² = 3.2/0.175

T = 4.28 s

|emf| = 4.0824x10⁻⁶ t  =>  4.0824x10⁻⁶ x4.28

|emf|= 1.74x10⁻⁵ V

Therefore,the magnitude of the emf induced in the secondary winding is 1.74x10⁻⁵ V

What can you infer about the prefix "thermo-" from the

word "thermoregulation?"

A

It refers to the brain

reser to the brain

It refers to temperature

It refers to feedback

It refers to nerves and hormones

Answers

It refers to temperature

Answer:

it refers to temperature I did test with the same question and got it correct.

Explanation:

Two loudspeakers are placed on a wall 2 m apart. A listener stands directly in front of one of the speakers, 81.7 m from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 4450 Hz. What is the phase difference ∆Φ between the two waves (generated by each speaker) when they reach the listener? The speed of the sound in air is 343 m/s. Answer in units of rad.

Answers

Answer:

The phase difference is       [tex]\Delta \phi = 1.9995 rad[/tex]  

Explanation:

From the question we are told that

    The distance between the  loudspeakers is [tex]d = 2m[/tex]

     The distance of the listener from the wall  [tex]D = 81.7 \ m[/tex]

     The frequency of the  loudspeakers is  [tex]f = 4450Hz[/tex]

      The velocity of sound is [tex]v_s = 343 m/s[/tex]

     

The path difference of the sound wave that is getting to the listener is mathematically represented as

        [tex]\Delta z =\sqrt{d^2 + D^2} -D[/tex]

Substituting values

        [tex]\Delta z =\sqrt{2^2 + 81.7^2 } -81.7[/tex]

       [tex]\Delta z =0.0245m[/tex]

The phase difference is mathematically represented as

           [tex]\Delta \phi[/tex] =  [tex]\frac{2 \pi}{\lambda } * \Delta z[/tex]

Where [tex]\lambda[/tex] is the wavelength which is mathematically represented as

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

substituting value  

          [tex]\lambda = \frac{343 }{4450}[/tex]

        [tex]\lambda = 0.0770 m[/tex]

Substituting value into the  equation for phase difference

      [tex]\Delta \phi[/tex] =  [tex]\frac{2 * 3.142 * 0.0245}{0.0770}[/tex]

      [tex]\Delta \phi = 1.9995 rad[/tex]  

You are packing for a trip to another star. During the journey, you will be traveling at 0.99c. You are trying to decide whether you should buy smaller sizes of your clothing because you will be thinner on your trip due to length contraction. You also plan to save money by reserving a smaller cabin to sleep in because you will be shorter when you lie down. Should you (a) buy smaller sizes of clothing, (b) reserve a smaller cabin, (c) do neither of these things, or (d) do both of these things?

Answers

Answer:

Do neither of these things ( c )

Explanation:

For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula

L = (l) [tex]\sqrt{1 -\frac{v^{2} }{c^{2} } }[/tex]

where l = Measured distance from object at rest, L =  contracted measured in relation to the observer , v = speed of clock , c = speed of light

you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .

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