An electric heater is rated at 1400 W, a toaster is rated at 1150 W, and an electric grill is rated at 1560 W. The three appliances are connected in parallel across a 112 V emf source. Find the current in the heater.

Answers

Answer 1

Answer:

The current in the heater is 12.5 A

Explanation:

It is given that,

Power of electric heater, P₁ = 1400 W

Power of toaster, P₂ = 1150 W

Power of electric grill, P₃ = 1560 W

All three appliances are connected in parallel across a 112 V emf source. We need to find the current in the heater. We know that in parallel combination of resistors the current flowing in every branch of resistor divides while the voltage is same.

Electric power, [tex]P_1=V\times I_1[/tex]

[tex]I_1=\dfrac{P_1}{V}[/tex]

[tex]I_1=\dfrac{1400\ W}{112\ V}[/tex]

[tex]I_1=12.5\ A[/tex]

So, the current in the heater is 12.5 A. Hence, this is the required solution.  


Related Questions

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the junction.What is the magnitude of the current in wire 3?

Answers

I₃ = 0.17 A into the junction.

The key to solve this problem is using Kirchhoff's Current Law which statements  that the algebraic sum of the currents that enter and leave a particular junction must be 0 (I₁+I₂+...+In = 0). Be careful, the currents that leaves a point is considered positive current, and the one that enters a point is considered negative.

In other words,  the sum of the currents that enter to the joint is equal to the sum of the currents that come out of the joint.

Three wire meet at junction. Wire 1 has a current of 0.40 A into the junction, the current of the wire 2 is 0.57 A out of the junction. What is the magnitude of the current in wire 3?

To calculate the current in the wire 3:

First, let's name the currents in the circuit. So:

The current in the wire 1 is I₁ = 0.40 A, the current in the wire 2 is I₂ = 0.57 A, and the current of wire 3 is I₃ = ?.

Second,  we make a scheme of the circuit, with the current I₁ of wire 1 into the junction, the current I₂ of wire 2 out of the junction, and let's suppose that the current I₃ of the wire 3 into the junction. (See the image attached)

Using Kirchhoff's Current Law, the sum of the currents I₁ and I₃ into the junction is equal to the current I₂ out of junction.

I₁ + I₃ = I₂

0.40 A + I₃ = 0.57 A

I₃ = 0.57 A - 0.40A

I₃ = 0.17 A

Checking the Kirchhoff's Current Law, the sum of all currents is equal to 0 :

I₁ + I₃ = I₂

I₁ - I₂ + I₃ = 0

0.40 A - 0.57 A + 0.17 A = 0

Let's suppose that the current I₃ of the wire 3 out the junction.

I₁ = I₂ + I₃

0.40 A = 0.57 A + I₃

I₃ = 0.40 A - 0.57 A

I₃ = -0.17 A which means that the current flow  in the opposite direction that we selected.

Final answer:

The magnitude of the current in wire 3 entering the junction is 0.17A. This result is based on the principle of conservation of electric current which states that the total current entering the junction must equal the total current exiting the junction.

Explanation:

This question involves the principle of the conservation of electric current, also known as Kirchhoff's Current Law. According to this concept, at any junction point the total current entering the junction is equal to the total current leaving the junction. In this case, if wire 1 has a current of 0.4A entering the junction and wire 2 has a current of 0.57A leaving the junction, the current in wire 3 would be the difference between these two values.

Since current can't be negative, if the current in wire 1 (entering the junction) is less than the current in wire 2 (exiting), the current in wire 3 must be entering the junction to balance the current flow.

Therefore, the magnitude of the current in wire 3 would be 0.57A - 0.4A = 0.17A. This means wire 3 has a current of 0.17A entering the junction.


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A particle moves along a circle with radius R, so that the tangential component of its acceleration is constant. At t = 0 the velocity of the particle was equal to zero. Find
(a) the normal component an of the acceleration as function of time.
(b) the magnitude of the acceleration vector a as well as the angle the vector a forms with the position vector r as functions of time.

Answers

Final answer:

To find the normal component of acceleration, differentiate the velocity and acceleration functions.

Explanation:

In order to find the normal component of the acceleration, we need to first find the velocity and acceleration vectors as functions of time. Using the given position function, we can differentiate it to find the velocity function:

v(t) = dr/dt = -3(4sin(3t)i + 4cos(3t)ĵ)

Next, we can differentiate the velocity function to find the acceleration function:

a(t) = dv/dt = -3(3cos(3t)i - 3sin(3t)ĵ)

Since the tangential component of the acceleration is constant, we can set the coefficient of the tangential component equal to a constant value:

-3(3sin(3t)) = Constant

Solving this equation will give us the value of the constant. Once we know the value of the constant, we can substitute it back into the acceleration function to find the normal component of the acceleration as a function of time.

A 26 tooth helical gear having a pitch diameter of 7 has a helix angle of 22 deg. Find the circular pitch in a plane normal to the teeth.

Answers

Answer:

the answer is that u need to divide the 26 by the 22 and then u add seven to that number and then u have to multiply by 2. And you will get your answer

Explanation:

A force of 34N stretches a very light ideal spring 0.73 m from equilibrium, What is the force constant (spring constant) of the spring? (A) 47N/m (B) 38N/m (C) 53N/m (D) 25N/m

Answers

Answer:

Spring constant, k = 47 N/m

Explanation:

It is given that,

Force applied to a spring, F = 34 N

A very light ideal spring moves 0.73 m from equilibrium position i.e. x = 0.73 m

We have to find the force constant or spring constant of the spring. It can be calculated using Hooke's law. According to him, the force acting on the spring when it compresses or stretches is given by :

[tex]F=-kx[/tex]  (-ve sign shows opposite direction)

[tex]k=\dfrac{F}{x}[/tex]

[tex]k=\dfrac{34\ N}{0.73\ m}[/tex]

k = 46.5 N/m

or

k = 47 N/m

Hence, the spring constant of the spring is 47 N/m.

Final answer:

Utilizing Hooke's law, which posits that the force needed to compress or extend a spring is proportional to the distance, the spring constant can be calculated to be approximately 46.6 N/m by dividing the force (34N) by the displacement (0.73m). So, the closest option amongst the provided ones is (A) 47N/m.

Explanation:

In the situation described where a force of 34N stretches a very light ideal spring 0.73 m from equilibrium, you would want to find the spring's force constant, often denoted as k. This involves a principle in physics known as Hooke's law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The equation for Hooke's law is F=kx.

To find the spring constant (k), you rearrange the equation for Hooke's Law to give you: k = F/x. Substituting the force (F = 34N) and displacement from equilibrium (x = 0.73m) into the equation gives: k = 34N / 0.73m. Calculating this gives a spring constant of approximately 46.6 N/m. So, none of the given options (A) 47N/m  (B) 38N/m  (C) 53N/m  (D) 25N/m are exactly correct, but Option A is the closest.

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A reaction taking place in a container with a piston-cylinder assembly at constant temperature produces a gas, and the volume increases from 127 mL to 654 mL against an external pressure of 860 torr. Calculate the work done in Joules (J)

Answers

Answer:

60.4 J

Explanation:

The work done by the gas is given by:

[tex]W=p(V_f-V_i)[/tex]

where

p is the gas pressure

[tex]V_f[/tex] is the final volume of the gas

[tex]V_i[/tex] is the initial volume

We must convert all the quantities into SI units:

[tex]p=860 torr \cdot \frac{1.013\cdot 10^5 pa}{760 torr}=1.146\cdot 10^5 Pa[/tex]

[tex]V_i = 127 mL = 0.127 L = 0.127 dm^3 = 0.127 \cdot 10^{-3}m^2[/tex]

[tex]V_f = 654 mL = 0.654 L = 0.654 dm^3 = 0.654 \cdot 10^{-3}m^2[/tex]

So the work done is

[tex]W=(1.146\cdot 10^5 Pa)(0.654\cdot 10^{-3} m^3-0.127\cdot 10^{-3} m^3)=60.4 J[/tex]

Answer:

[tex]W=60.4 J[/tex]

Work done is 60.4 Joules (J)

Explanation:

Work done 'W' is given by:

[tex]W=P\triangle V[/tex]

Where:

ΔV is the change in Volume.

P is the pressure.

Change in Volume=Final Volume-Initial Volume

Initial Volume= 127 mL

Final Volume= 654 mL

ΔV=654-127 (mL)

ΔV=527 mL

ΔV=0.527 L

Pressure Conversion:

1 atm=760 torr

[tex]P=\frac{860}{760} \\P=1.1315\ atm[/tex]

Now,

[tex]W=P\triangle V[/tex]

[tex]W=1.1315\ atm*0.527\ L[/tex]

[tex]W=0.5963 L.atm[/tex]

In joule (J): (Conversion 1 atm. L=101.325 J)

[tex]W=0.5963\ L.atm*\frac{101.325 J}{1\ L.atm}[/tex]

[tex]W=60.4 J[/tex]

Work done is 60.4 Joules (J)

True or False? The surfaces of equal potential around an infinite charged plane are parallel to the plane (i.e. the surfaces never intersect the charged plane). False True [5 points] Suppose I center a conducting sphere of charge (radius R, total charge +Q) at the origin. How much energy would it take to move a point charge (total charge +q) from the surface of the sphere (r = R) to the center of the sphere (r = 0)? E = kq / R

Answers

Answer:

true

Explanation:

the sun hits it ig idek

The surfaces of equal potential around an infinite charged plane are parallel to the plane: True.

The surfaces around an infinite charged plane.

According to the law of electrostatic forces, the surfaces of equal potential around an infinite charged plane are always parallel to the plane because they cannot intersect the charged plane.

Since this conducting sphere of charge is centered at the origin (0), the quantity of energy that would be required to move a point charge from the surface of this sphere (r = R) to the center of the sphere is given by:

E = qV

E = q × 0

E = 0 Joules.

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A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the car from sliding down the incline.

Answers

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

[tex]f=mg\ sin\theta[/tex]

[tex]f=1100\ kg\times 9.8\ m/s^2\ sin(4)[/tex]

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

Final answer:

In essence, the force of friction that prevents the car from sliding down the incline is equal to the component of the weight of the car down the incline. After performing the necessary calculations, this is found to be roughly 766.9 Newtons.

Explanation:

In response to your question, the force of friction keeping the car from sliding down the incline can be calculated using basic physics principles. Firstly, we need to calculate the component of the gravitational force parallel to the incline, which is given by F_gravity = m * g * sin(theta), where m is the mass of the car, g is the acceleration due to gravity, and theta is the angle of incline.

Substituting the given values, we get F_gravity = 1100 kg * 9.8 m/s² * sin(4°), which approximates to 766.9 N.

Now, since the car is stationary and not sliding down, this suggests that the force of friction equals the component of the weight down the incline. Therefore, the force of friction keeping the car from sliding down is approximately 766.9 Newtons.

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A voltage source Vs = 10 V is in series with a resistor of 10 kOhm. If the source transformation theorem is applied, what will the value of the current source be in mA ?

Answers

Answer:

The current source is 1 mA.

Explanation:

Given that,

Voltage = 10 V

Resistor = 10 k ohm

We need to calculate the current

Using formula of transformation theorem

The source transformation must be constrained is given by

[tex]v_{s}=iR_{s}[/tex]

Where, V = voltage

I = current

R = resistor

Put the value into the formula

[tex]10=i\times10\times10^{3}[/tex]

[tex]i =\dfrac{10}{10\times10^{3}}[/tex]

[tex]i = 0.001\ A[/tex]

[tex]i=1\ mA[/tex]

Hence, The current source is 1 mA.

Final answer:

By using Ohm's Law (I = V/R), it can be determined that for a voltage source Vs = 10 V in series with a resistor of 10 kOhm, the current source would be 1mA after source transformation.

Explanation:

The question asks for the current value in a circuit when the source transformation theorem is applied. This involves using Ohm's law, which in basic form states that the current through a resistor equals the voltage across the resistor divided by the resistance. It's represented by the equation I = V/R.

In our case, the voltage (V) is 10 volts, and the resistance (R) is 10 kilo-Ohms or 10,000 Ohms. Applying Ohm's law gives I = V/R, so I = 10V/10,000Ω = 0.001 A. However, because the question asks for the current in milliamperes (mA), we need to convert from amperes to milliamperes, knowing that 1A = 1000mA. Therefore, 0.001A = 1mA.

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Two forces act on a 55-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other has a magnitude 35 N at 32° clockwise from the positive y-axis. What is the magnitude of this objectʹs acceleration? A) 1.1 m/s2 B) 1.3 m/s2 C) 1.5 m/s2

Answers

Answer:

A) 1.1 m/s/s

Explanation:

There exist two forces on the object such that

[tex]F_1[/tex] = 65 N directed 59° clockwise from the positive x-axis

[tex]F_2[/tex] = 35 N at 32° clockwise from the positive y-axis

now we have

[tex]F_1 = 65 cos59\hat i - 65 sin59 \hat j[/tex]

[tex]F_2 = 35 sin32\hat i + 35 cos32 \hat j[/tex]

now the net force on the object is given as

[tex]F_{net} = F_1 + F_2[/tex]

[tex]F_{net} = (65 cos59 + 35 sin32)\hat i + (35cos32 - 65 sin59)\hat j[/tex]

[tex]F_{net} = 52\hat i - 26 \hat j[/tex]

so it's magnitude is given as

[tex]F_{net} = \sqrt{52^2 + 26^2} = 58.15 N[/tex]

now from Newton's II law we have

F = ma

[tex]a = \frac{58.15}{55} = 1.1 m/s^2[/tex]

Final answer:

To find the magnitude of the object's acceleration, resolve the forces into components and use Newton's second law and the Pythagorean theorem.

Explanation:

To find the magnitude of the object's acceleration, we need to first resolve the two forces into their x and y components. For the 65 N force, the x-component is 65 N × cos(59°) and the y-component is 65 N × sin(59°). For the 35 N force, the x-component is 35 N × sin(32°) and the y-component is 35 N × cos(32°).

Next, we add the x-components and the y-components separately to get the net force in the x and y directions. Then we use Newton's second law, F = ma, to find the acceleration in each direction. Finally, we use the Pythagorean theorem to find the magnitude of the acceleration.

By following these steps, we can calculate that the magnitude of the object's acceleration is 1.5 m/s², so the correct answer is C) 1.5 m/s².

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Suppose you dissolve 4.75 g of another solid (not urea) in 50.0 mL of water. You note that the temperature changes from 25.0 °C to 28.0 °C. Part 1: What is the mass of the solution?

Answers

Answer:

Mass of the solution  = 54.75 g

Explanation:

Mass of solid dissolved = 4.75 g

Mass of the solution = Mass of solid dissolved + Mass of water.

Mass of water = Volume x Density

Volume = 50 mL = 50 cm³

Density = 1 g/cm³

Mass of water = 50 x 1 = 50 g

Mass of the solution = 4.75 + 50 = 54.75 g

An ideal transformer has 60 turns in its primary coil and 360 turns in its secondary coil. If the input rms voltage for the 60-turn coil is 120 V, what is the output rms voltage of the secondary coil?

Answers

Answer:

Voltage of the secondary coil is 720 volts.

Explanation:

Number of turns in primary coil, N₁ = 60

Number of turns in secondary coil, N₂ = 360

Input rms voltage of primary coil, V₁ = 120 V

We have to find the output rms voltage of the secondary coil. The relationship between the number of coil and voltage is given by :

[tex]\dfrac{N_1}{N_2}=\dfrac{V_1}{V_2}[/tex]

[tex]\dfrac{60}{360}=\dfrac{120}{V_2}[/tex]

V₂ = output rms voltage of the secondary coil.

After solving above equation, we get :

V₂ = 720 V

Hence, the output rms voltage of the secondary coil is 720 volts.

What will happen to the volume of a gas if its absolute temperature triples (increases by a factor of three) as the amount of gas and the pressure are held constant? Select the correct answer below: Question 3 options: The volume will decrease by a factor of three. The volume will triple (increase by a factor of three). The volume will double (increase by a factor of two).

Answers

Answer: The volume will triple (increase by a factor of three).

Explanation:

The expression for an Ideal Gas is:  

[tex]P.V=n.R.T[/tex]    (1)

Where:  

[tex]P[/tex] is the pressure of the gas  

[tex]V[/tex] is the volume of the gas  

[tex]n[/tex] the number of moles of gas  

[tex]R[/tex] is the gas constant  

[tex]T[/tex] is the absolute temperature of the gas  

Finding [tex]V[/tex]:

[tex]V=\frac{n.R.T}{P}[/tex]    (2)

If we are told the the amount of gas [tex]n[/tex] and pressure [tex]P[/tex] remain constant, but we increase the temperature [tex]T[/tex] by a factor of three; we will have to rewrite (2) with the new temperature [tex]T_{N}[/tex]:

[tex]T_{N}=3T[/tex]

[tex]V=\frac{n.R.3T}{P}[/tex]    (3)

[tex]V=3\frac{n.R.T}{P}[/tex]    (4)

Now, if we compare (2) with (4), it is clearly noticeable the volume of the gas has increased by a factor of 3.

The volume will triple ( increase by a factor of three )

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Further explanation

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

[tex]\large {\boxed {P = F \div A} }[/tex]

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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In this problem , we will use Ideal Gas Law as follows:

Given:

Initial Temperature of The Gas = T₁

Final Temperature of The Gas = T₂ = 3T₁

Asked:

Final Volume of The Gas = V₂ = ?

Solution:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]\frac{PV_1}{T_1} = \frac{PV_2}{3T_1}[/tex]

[tex]\frac{V_1}{1} = \frac{V_2}{3}[/tex]

[tex]3(V_1) = 1(V_2)[/tex]

[tex]V_2 = 3V_1[/tex]

[tex]\texttt{ }[/tex]

Conclusion:

The volume will triple ( increase by a factor of three )

[tex]\texttt{ }[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure, Volume , Ideal , Gas , Law

At a distance of 20 m from a source of sound, the sound level is 40 dB. If the observer backs up to a distance of 40 m from the source, what will the sound level be? O A. 34 dB O B. 28 dB ° C. 20 dB O D. 10 dB

Answers

Answer:

The sound level is 34 dB.

(A) is correct option.

Explanation:

Given that,

Distance from a source of sound [tex]r_{1}=20 m[/tex]

Sound level [tex]L_{1}= 40\ dB[/tex]

Distance from a source of sound[tex]r_{2}=40\ m[/tex]

We need to calculate the sound level

Using equation

[tex]L_{2}=L_{1}-|20 log(\dfrac{r_{1}}{r_{2}})|[/tex]

[tex]L_{2}=40-|20 log(\dfrac{20}{40})|[/tex]

[tex]L_{2}=40-|20 log(0.5)|[/tex]

[tex]L_{2}=33.9\ dB[/tex]

[tex]L_{2}=34\ dB[/tex]

Hence, The sound level is 34 dB.

The sound level decreases by 20 dB when the distance from the source is doubled because the intensity of sound decreases with the square of the distance. Thus, the sound level at 40 m would be 20 dB. The answer is option C.

When the observer moves from a distance of 20 m to 40 m away from the source of sound, the sound intensity decreases. This is due to the inverse square law, which states that the intensity of sound is inversely proportional to the square of the distance from the source. Since the observer doubles the distance from the source, the intensity of the sound will be reduced to one-fourth. Because each 10-fold decrease in intensity corresponds to a reduction of 10 dB in sound level, halving the distance corresponds to an intensity ratio of 1/4, which is two factors of 10 (or 102 times less intense). Thus, the sound level decreases by 20 dB (2 factors of 10).

Therefore, at a distance of 40 m, the sound level would be 20 dB less than the original 40 dB, leading to a new sound level of 20 dB. The answer is option C: 20 dB.

The emf of a battery is equal to its terminal potential difference A) under all conditions B) only when the battery is being charged C) only when a large current is in the battery D) only when there is no current in the battery E) under no conditions

Answers

D. The chemical potential energy of real battery decreases, it develops internal resistance and therefore the potential difference across its terminals decreases if its current increases.

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round the bend. The radius of the curve is 140 m. Compute the acceleration at the moment the train speed reaches 46.0 km/h. Assume the train continues to slow down at this time at the same rate.

Answers

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

[tex]u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s[/tex]

[tex]v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s[/tex]

Now we find the centripetal acceleration which is given by

[tex]a_c=\frac{v^2}{r}[/tex]

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

[tex]a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2[/tex]

we also have a tangential acceleration, which is given by

[tex]a_t = \frac{v-u}{t}[/tex]

where

t = 17.0 s

Substituting values,

[tex]a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2[/tex]

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

[tex]a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2[/tex]

Answer:

a = 1.406 m/s²

Explanation:

We are told that, the speed of the train decreases from 94.0 km/h to 46.0 km/h

Let's convert both to m/s.

Thus,

v1 =94 km/h =(94 x 10)/36 =26.11 m/s

v2=46 km/h =(46 x 10)/36=12.78 m/s

The formula to calculate the tangential acceleration is given by;

a_t = -dv/dt

Where;

dv is change in velocity

dt is time difference

dv is calculated as; dv = v1 - v2

Thus, dv = 26.11 - 12.78 = 13.33 m/s

We are given that, t = 17 seconds

Thus;

a_t = -13.33/17 = -0.784 m/s²

The negative sign implies that the acceleration is inwards.

Now, let's calculate the radial acceleration;

a_r = v²/r

Where;

r is the radius of the path = 140m

v is the velocity at the instant given

a_r is radial acceleration.

Thus,

a_r = 12.78²/140 = 1.167 m/s²

Now, the tangential and radial components of acceleration are perpendicular to each other. Thus, we can use using Pythagoreas theorem to find the resultant acceleration;

Thus;

a² = (a_t)² + (a_r)²

Plugging in the relevant values, we have;

a² = (-0.784)² + (1.167)²

a² = 1.976545

a = √1.976545

a = 1.406 m/s²

Suppose that an electromagnetic wave which is linearly polarized along the x−axis is propagating in vacuum along the z−axis. The wave is incident on a conductor which is placed at z > 0 region of the space. The conductor has conductivity σ, magnetic permeability µ and electric permittivity ε.
(a) Find the characteristic time for the free charge density which dissipates at the conductor.

(b) Write the Maxwell equations and derive the wave equation for a plane wave propagating in a conductor.

(c) Find the attenuation distance at which the incident amplitude reduces to e ^−1 of its initial value.

(d) Find the electric and magnetic fields inside the conductor. 8 (e) Find the power loss per area of the incident electromagnetic wave at the surface of conductor.

Answers

I think the answer to your question is B

Romeo (75.1 kg) entertains Juliet (58.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek. How far does the 76.5-kg boat move toward the shore it is facing?

Answers

Answer: I think 8.7 m

Explanation:

A bullet with a mass ????=12.5 g and speed ????=86.4 m/s is fired into a wooden block with ????=113 g which is initially at rest on a horizontal surface. The bullet is embedded into the block. The block-bullet combination slides across the surface for a distance ???? before stopping due to friction between the block and surface. The coefficients of friction are ????????=0.753 and ????????=0.659. (a) What is the speed (m/s) of the block-bullet combination immediately after the collision? (b) What is the distance d (m)?

Answers

Answer:

a) 8.61 m/s, b) 5.73 m

Explanation:

a) During the collision, momentum is conserved.

mv = (m + M) V

(12.5 g) (86.4 m/s) = (12.5 g + 113 g) V

V = 8.61 m/s

b) After the collision, energy is conserved.

Kinetic energy = Work done by friction

1/2 (m + M) V² = F d

1/2 (m + M) V² = N μk d

1/2 (m + M) V² = (m + M) g μk d

1/2 V² = g μk d

d = V² / (2g μk)

d = (8.61 m/s)² / (2 × 9.8 m/s² × 0.659)

d = 5.73 m

Notice we used the kinetic coefficient of friction.  That's the friction when an object is moving.  The static coefficient of friction is the friction on a stationary object.  Since the bullet/block combination is sliding across the surface, we use the kinetic coefficient.

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of 1.43 cm2 and the woman's mass is 60.5 kg. Express the pressure in pascals. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.)

Answers

Answer:

4.15 x 10^6 N

Explanation:

Area, A = 1.43 cm^2 = 1.43 x 10^-4 m^2

mass, m = 60.5 kg

Weight, F = m g = 60.5 x 9.8 = 592.9 N

Pressure = Force / Area

P = Weight / Area

P = 592.9 / (1.43 x 10^-4)

P = 4.15 x 10^6 N

In an amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is 3.2 m above the bottom where they exit the slide, moving horizontally, 1.2 m above a swimming pool. Does the mass of the person make any difference?

Answers

Answer:

No

Explanation:

When the person slides down, the change in gravitational potential energy is converted into kinetic energy, according to

[tex]\Delta U = \Delta K\\mg\Delta h = \frac{1}{2}mv^2[/tex]

where

m is the mass of the person

g is the acceleration of gravity

v is the final speed

[tex]\Delta h[/tex] is the change in heigth of the person

Here we have assumed that the initial speed is zero.

Re-arranging the equation,

[tex]v = \sqrt{2g \Delta h}[/tex]

and we see that this quantity does not depend on the mass of the person, so every person will have the same speed at the bottom of the slide, equal to:

[tex]v=\sqrt{2(9.8 m/s^2)(3.2 m-1.2 m)}=6.3 m/s[/tex]

Compute the voltage drop along a 24-m length of household no. 14 copper wire (used in 15-A circuits). The wire has diameter 1.628 mm and carries a 13-A current. The resistivity of copper is 1.68×10−8Ω⋅m.

Answers

Answer:

Explanation:

We know that the formula for resistance in terms of length and area is given by

R = p l/ a

Where p be the resistivity and a be the area of crossection

a = pi × d^2 / 4

Where d be the diameter

a = 3.14 × (1.628 × 10^-3)^2 / 4

a = 2 × 10^-6 m^2

R = 1.68 × 10^-8 × 24 / (2 × 10^-6)

R = 20.16 × 10^-2 ohm

By use of Ohm's law

V = IR

V = 13 × 20.16 × 10^-2

V = 2.62 Volt

How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad/s^2?

Answers

Answer:

2.17 s

Explanation:

Here, w0 = 33.3 rpm = 33.3 /60 rps

= 2 × pi × 33.3 / 60 rad/ s

= 3.4854 rad / s

w = 78 rpm = 78 / 60 rps

= 2 × pi × 78 / 60 = 8.164 rad / s

a = 2.15 rad/s^2

Use first equation of motion for rotational motion

w = w0 + a t

8.164 - 3.485 = 2.15 × t

t = 2.17 s

A highway curve with radius 80 m is to be banked so that a car traveling 21 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?

Answers

Answer:

Angle, θ = 29.35

Explanation:

It is given that,

Radius of the curve, r = 80 m

Velocity of the car, v = 21 m/s

The highway curve is to be banked so that the car will not skid sideways in the presence of friction. We need to find the angle should the curve be banked. Let the angle is θ. It is given by :

[tex]tan{\theta}=\dfrac{v^2}{rg}[/tex]

Where

g = acceleration due to gravity

[tex]tan\ {\theta}=\dfrac{(21\ m/s)^2}{80\ m\times 9.8\ m/s^2}[/tex]

[tex]\theta=tan^{-1}(0.5625)[/tex]

θ = 29.35°

Hence, this is the required solution.

A circular coil of radius 10 cm and a separate square coil of side 20 cm are both rotated in a magnetic field of 1.5 T. If the circular coil is rotated at a frequency of 60 Hz, then at what frequency must the square coil be rotated in order for both coils to have the same maximum induced voltage? A) 47 Hz
B) 60 Hz
C) 76 Hz
D) 19 Hz

Answers

Answer:

The frequency of square coil is 47 Hz.

(A) is correct option.

Explanation:

Given that,

Radius =10 cm

Side = 20 cm

Magnetic field = 1.5 T

Frequency = 60 Hz

We need to calculate the the maximum induced voltage

[tex]V_{m}=BAN\omega[/tex]

Where, B = magnetic field

A = area of cross section

N = number of turns

[tex]\omega=2\pi f[/tex]= angular frequency

Put the value into the formula

[tex]V_{m}=1.5\times3.14\times(10\times10^{-2})^2\times1\times2\times3.14\times60[/tex]

[tex]V_{m}=17.75\ V[/tex]

If the square coil have the same induced voltage.

Area of square of coil [tex]A =(20\times10^{-2})^2[/tex]

[tex]A=0.04\ m^2[/tex]

Now, The angular velocity of square coil

[tex]\omega=\dfrac{V_{m}}{AB}[/tex]

[tex]\omega=\dfrac{17.75}{0.04\times1.5}[/tex]

[tex]\omega=295.8\ \dfrac{rad}{s}[/tex]

Now, frequency of rotation

[tex]f = \dfrac{\omega}{2\pi}[/tex]

Put the value into the formula of frequency

[tex]f=\dfrac{295.8}{2\times3.14}[/tex]

[tex]f=47\ Hz[/tex]

Hence, The frequency of square coil is 47 Hz.

A particle leaves the origin with an initial velocity of 3.00 m/s in the x direction, and moves with constant acceleration ax = -2.70 m/s2 and ay = 3.90 m/s2. How far does the particle move in the x direction before turning around?

Answers

Answer:

1.67 m

Explanation:

Ux = 3 m/s, ax = -2.7 m/s^2, vx =0

Let the distance travelled before stopping along x axis is x.

Use third equation of motion along x axis

Vx^2 = ux^2 + 2 a x

0 = 9 - 2 × 2.7 × x

X = 1.67 m

initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m/s at an angle of 54.2° below the horizontal. What was the particle's average acceleration during these 2.00 seconds in the x-direction (enter first) and the y-direction?

Answers

Explanation:

Average acceleration is the change in velocity over the change in time:

a = (v − v₀) / t

In the x direction:

aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00

aₓ = -0.316 m/s²

In the y direction:

aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00

aᵧ = -4.11 m/s²

Final answer:

To find the average acceleration, decompose the given velocities into their x and y components. Then calculate the change in velocities in both directions divided by the time interval.

Explanation:

The average acceleration of a particle can be found using the change in velocity and the time interval. The given velocities are in magnitude and direction, so we need to decompose these into their x and y components. The initial and final x-components are Vix = 5.33 m/s cos(37.9°) and Vfx = 6.11 m/s cos(54.2°), respectively. The average x-direction acceleration (ax) can be calculated as ax = (Vfx - Vix)/time. Similarly, the initial and final y-components are Viy = 5.33 m/s sin(37.9°) and Vfy = 6.11 m/s sin(54.2°), respectively. The y-direction acceleration (ay) is ay = (Vfy - Viy)/time.

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A parallel-plate capacitor has a plate area of 0.3 m2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 5 × 10–6 C then the force exerted by one plate on the other has a magnitude of about:

Answers

Answer:

4.72 N

Explanation:

The charge density across each plate is given by:

[tex]\sigma = \frac{Q}{A}[/tex]

where

[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate

[tex]A=0.3 m^2[/tex] is the area of each plate

Solving,

[tex]sigma = \frac{5\cdot 10^{-6}C}{0.3 m^2}=1.67\cdot 10^{-5} C/m^2[/tex]

The force exerted by one plate on the other is given by:

[tex]F=\frac{Q\sigma}{2\epsilon_0}[/tex]

where

[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate

[tex]\sigma=1.67\cdot 10^{-5} C/m^2[/tex] is the surface charge density

[tex]\epsilon_0[/tex] is the vacuum permittivity

Substituting,

[tex]F=\frac{(5\cdot 10^{-6} C)(1.67\cdot 10^{-5} C/m^2)}{2(8.85\cdot 10^{-12}F/m)}=4.72 N[/tex]

Final answer:

We can find the force exerted by one plate of a parallel-plate capacitor on the other by utilizing the known quantities and the formula F = Q² / (A * ε), which arises from force definition and electric field for a parallel-plate capacitor.

Explanation:

The magnitude of the force between two charges is given by Coulomb's law: F = k * (Q1 * Q2) / d², where k is the Coulomb's constant, Q1 and Q2 are the charges, and d is the distance. The electric field for a parallel-plate capacitor is E = Q / (A * ε), where ε is the permittivity of free space.

The force exerted on each plate of the capacitor is F = QE which, once substituted, becomes F = Q² / (A * ε) when charge, plate area, and permittivity are known. Given the values of the problem, by substituting into the equation, you can calculate the force between the plates.

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As mentioned before, our asteroid is in the shape of a sphere and has a mass of 1000 kilograms. Determine the density (in grams per cubic centimeter) of this asteroid if its diameter is known to be 1.2 meters. Useful information: 1 kg = 1000 g, 1 m = 100 cm, volume of sphere = 4/3 ? r3. Remember that the radius of a sphere is equal to half its diameter. Show all of your work. (20 points)

Answers

Answer: [tex]1.1052g/cm^{3}[/tex]

Explanation:

Density [tex]D[/tex] is a characteristic property of a material and is defined as the relationship between the mass [tex]m[/tex] and volume [tex]V[/tex] of a specific substance or material. So, the density of the asteroid is given by the following equation:

[tex]D=\frac{m}{V}[/tex]   (1)

On the other hand, we know the asteroid has a mass [tex]m=1000kg[/tex] and is spherical. This means its volume is given by the following formula:

[tex]V=\frac{4}{3}}\pi r^{3}[/tex]   (2)

Where [tex]r=\frac{d}{2}=\frac{1.2m}{2}=0.6m[/tex]  is the radius of the sphere and is half its diameter [tex]d[/tex].

Knowing this, we can calculate the volume:

[tex]V=\frac{4}{3}}\pi (0.6m)^{3}[/tex]   (3)

[tex]V=0.904m^{3}[/tex]   (4)

Substituting (4) in (1):

[tex]D=\frac{1000kg}{0.904m^{3}}=1105.242\frac{kg}{m^{3}}[/tex]   (5) This is the density of the asteroid, but we were asked to find it in [tex]\frac{g}{cm^{3}}[/tex]. This means we have to make the conversion:

[tex]D=1105.242\frac{kg}{m^{3}}.\frac{1000g}{1kg}.\frac{1m^{3}}{(100cm)^{3}}[/tex]

Finally:

[tex]D=1.1052\frac{g}{cm^{3}}[/tex]

A solenoid 0.425 m long has 950 turns of wire. What is the magnetic field in the center of the solenoid when it carries a current of 2.75 A? (JC 19.57)

Answers

Answer:

The magnetic field in the center of the solenoid is [tex]7.8\times10^{-3}\ T[/tex].

Explanation:

Given that,

Length of solenoid = 0.425 m

Number of turns N = 950

Current I = 2.75 A

The magnetic field in the center of the solenoid is the product of the current , number of turns per unit length and permeability.

In mathematical form,

[tex]B = \mu_{0}nI[/tex]

Where, [tex]n = \dfrac{N}{l}[/tex]

N = number of turns

L = length

I = current

Now, The magnetic field

[tex]B = \dfrac{\mu_{0}NI}{l}[/tex]

Put the value into the formula

[tex]B=\dfrac{4\pi\times10^{-7}\times950\times2.75}{0.425}[/tex]

[tex]B=\dfrac{4\times3.14\times10^{-7}\times950\times2.75}{0.425}[/tex]

[tex]B=7.8\times10^{-3}\ T[/tex]

Hence, The magnetic field in the center of the solenoid is [tex]7.8\times10^{-3}\ T[/tex].

At what height above the ocean the acceleration of gravity of the earth will have a value of 9.3m /s^2 ?

Answers

Answer:

160 km

Explanation:

g = 9.8 m/s^2, g' = 9.3 m/s^2

Let h be the height from ocean level.

Use the formula for acceleration due to gravity at height.

g' = g (1 - 2h/R)

The radius of earth is 6400 km

9.3 = 9.8 ( 1 - 2 H / R)

0.05 = 2 h/R

h = 0.05 × 6400 / 2 = 160 km

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