How many loops are in this circuit?

How Many Loops Are In This Circuit?

Answers

Answer 1
I think there a 4 but I am not sure
Answer 2

I see six (6) loops.

I attached a drawing to show where I get six loops from.

How Many Loops Are In This Circuit?

Related Questions

A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg-m2
and is rotating at 10 rev/min. A 25-kilogram child at rest jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?

Answers

Answer:

dont be lose because the person who lose will win the match

Why are objects measured?​

Answers

In order to find out how long/wide/heavy/high/dense/deep/ massive/voluminous/reflective/opaque/ tansparent/warm/cold/hard/soft/ malleable/flexible/rigid/radioactive/old/ valuable/symmetrical/flat/regular/ irregular they are.

In a way that you can easily and conveniently describe to other people.


If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?

Answers

Answer:

the current that flows through the component is 2.42 A

Explanation:

Given;

resistance of the electrical component, r = 53 Ω

the voltage of the source, V = 128 V

The current that flows through the component is calculated using Ohm's Law as demonstrated below;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]

Therefore, the current that flows through the component is 2.42 A

You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.

Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?

Answers

Hi there!

a.

Use the formula d = st to solve:

d = 20 × 0.5 = 10m

150 - 10 = 140m away when brakes are applied

b.

Use the following kinematic equation to solve:

vf² = vi² + 2ad

Plug in known values:

0 = 20² + 2(150)(a)

Solve:

0 = 400 + 300a

-300a = 400

a = -4/3 (≈ -1.33) m/s² required

c.

Use the following kinematic equation to solve:

vf = vi + at

0 = 20 - 4/3t

Solve:

4/3t = 20

Multiply both sides by 3/4 for ease of solving:

t = 15 sec

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answers

Answer:

a n c

Explanation:

why is it wrong to leave our light on​

Answers

Answer:

you will get huge electricity bills ............

3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive

Answers

Answer:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

Explanation:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.

After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.

I need help with this physics question.

Answers

Answer:

5.04 m

Explanation:

You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.

Answer:

5.0384m

Explanation:

% increase = 100 x (Final - Initial / | initial | )

( |~~| Bars indicate absolute value since you can't have a negative height)

A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds

Answers

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Answers

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Workdone

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

W=4.49*10e10 J

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Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.

Answers

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.

Answers

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor

Answers

Answer:

560.54 Ω

Explanation:

Applying,

V = IR'............... Equation 1

Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors

make R' the subject of the equation

R' = V/I............ Equation 2

From the question,

Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A

Substitute these values into equation 2

R' = 5/(2.23×10⁻³ )

R' = 2242.15 Ω

Since the fours resistor are connected in series and they are equal,

Therefore the values of each resistor is

R = R'/4

R = 2242.15/4

R = 560.54 Ω

A car of mass 500 kg increases its velocity from 40 metre per second to 60 metre per second in 10 second find the distance travelled and amount of force applied ​

Answers

Answer:

it is answer of u are question

An airplane increases its speed at the average rate of 15 m/s2. How much time does it take to increase its speed from 100 m/s to 160 m/s

Answers

Answer:

4 s

Explanation:

Acceleration (a) = 15 m/s²Initial velocity (u) = 100 m/sFinal velocity (v) = 160 m/s

We are asked to calculate time taken (t).

By using the first equation of motion,

[tex]\longrightarrow[/tex] v = u + at

[tex]\longrightarrow[/tex] 160 = 100 + 15t

[tex]\longrightarrow[/tex] 160 - 100 = 15t

[tex]\longrightarrow[/tex] 60 = 15t

[tex]\longrightarrow[/tex] 60 ÷ 15 = t

[tex]\longrightarrow[/tex] 4 s = t

What are the differences among elements, compounds, and mixtures?

Answers

Answer:

Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.

••••••••••••••••

Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.

•••••••••••••••

A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements. 

------------------------------

Hope it helps...

Have a great day!!!

Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.

Which of the following statements is correct about the magnitude of the static friction force between an object and a surface?

a. Static friction depends on the mass of the object.
b. Static friction depends on the shape of the object.
c. Static friction depends on what the object is made of but not what the surface is made of.
d. None of the above is correct.

Answers

Answer:

Static friction depends on the mass of the object.

Explanation:

Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;

1) nature of the object and the surface(how rough or smooth the surfaces are)

2)surface area of the object and the surface

3) mass of the object

Since;

F=μmg

Where;

μ= coefficient of static friction

m= mass of the object

g= acceleration due to gravity

Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.

The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?

Answers

Answer:

Acosθ

Explanation:

The x-component of a vector is defined as :

Magnitude * cosine of the angle

Maginitude * cosθ

The magnitude is represented as A

Hence, horizontal, x - component of the vector is :

Acosθ

Furthermore,

The y-component is taken as the sin of the of the angle multiplied by the magnitude

Vertical, y component : Asinθ

The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?

Answers

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces an amount of electric energy with the hot reservoir at 500 K during Day One and then produces the same amount of electric energy with the hot reservoir at 600 K during Day Two. The thermal pollution was:

Answers

Answer: hello your question lacks some vital information below is the complete question

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was

answer:

Total thermal pollution = 2.5 * 10^6 J

Explanation:

Low temperature reservoir = 300 K

hot reservoir temperature = 500 K

Electrical energy produced by plant ( W ) = 1 * 10^6 J

lets assume ; Q1 = energy absorbed , Q2 = energy emitted

W = Q1 - Q2  or  Q2 = Q1 - W  ( we will apply this as the formula for determining thermal pollution )

For day 1

T1 = 500k , T2 = 300k

applying Carnot engine formula

W / Q1 = 1 - T2/T1

∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J

thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J

for Day 2

T1 = 600k,  T2 = 300k

Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J

Thermal pollution; Q2 = Q1 - W  = 1 * 10^6 J

Therefore the Total thermal pollution =  1 * 10^6  + 1.5 * 10^6  = 2.5 * 10^6 J

How are elastic and inelastic collisions different?


A: Elastic collisions occur when the colliding objects move separately after the collision; after inelastic collisions, the objects are connected and move together.

B: Elastic collisions occur when the objects are going the same direction when they collide; inelastic collisions occur when the objects are going in opposite directions when they collide.

C: Momentum is conserved in elastic collisions; momentum is not conserved in inelastic collisions.

D: Elastic collisions occur between objects of the same mass; inelastic collisions occur between different masses.

Answers

Answer:

a

Explanation:

Answer:

the answer is c

'

Explanation:

A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)

Answers

Answer:

Average Velocity = 3.63 m/s

Explanation:

First, we will calculate the total displacement of the quarterback, taking forward direction as positive:

Total Displacement = 15 m - 3 m + 24 m = 36 m

Now, we will calculate the total time taken for this displacement:

Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s

Therefore, the average velocity will be:

[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]

Average Velocity = 3.63 m/s

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration

Answers

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]

Therefore, the frequency of this mode of vibration is 138.87 Hz

convert 2.4 milimetres into metre​

Answers

0.0024

Milimetres are before centimetres and centimetres are before metres

Answer: 2.4 millimeters = 0.0024 meters

Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.

A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the container the pressure is 116 kPa.
(a) What is the depth of the fluid?
(b) Find the pressure at the bottom of the container after an additional 2.05 X 10^-3 m^3 of this fluid is added to the container. Assume that no fluid spills out of the container.

Answers

The right answer is (b)

A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q

Answers

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N

Coulomb's law equation

F = Kq₁q₂ / r²

Where

F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart

Data obtained from the question Initial distance apart (r₁) =  rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =?

How to determine the final force

From Coulomb's law,

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ = constant

F₁r₁² = F₂r₂²

With the above formula, we can obtain the final force as follow:

F₁r₁² = F₂r₂²

1.2 × r² = F₂ × (2r)²

1.2r² = F₂ × 4r²

Divide both side by 4r²

F₂ = 1.2r² / 4r²

F₂ = 0.3 N

Learn more about Coulomb's law:

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A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.

Answers

B it’s Intail velocity

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval

Answers

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

       

I hope it helps you!

g four small masses 0.2 kg each are connected by light rods 0.4m long to form a square.what is the moment of interia axis

Answers

Complete Question

Four small masses of 0.2 kg each are connected by light rods 0.4m long to form a square. What is the moment of inertia of this object for an axis through the middle of the square and parallel to two sides.

Answer:

[tex]I=0.032kgm^2[/tex]

Explanation:

From the question we are told that:

Mass[tex]m=0.2kg[/tex]

Length [tex]l=0.4m[/tex]

Generally the equation for Inertia is mathematically given by

 [tex]I=md^2[/tex]

 [tex]I=0.8*0.20(\frac{0.40}{2})^2[/tex]

 [tex]I=0.032kgm^2[/tex]

Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2.
(a) When a constant force is applied to object 1, it accelerates through a distance d. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (Select all that apply.)
K1 < K2 p1 = p2 p1 < p2 p1 > p2 K1 > K2 K1 = K2
(b) When a force is applied to object 1, it accelerates for a time interval ?t. The force is removed from object 1 and is applied to object 2. Which statements are true after object 2 has accelerated for the same time interval ?t? (Select all that apply.)
K1 > K2 K1 = K2 p1 = p2 p1 > p2 K1 < K2 p1 < p2

Answers

Answer:

Look at explanation

Explanation:

a) Kinetic energy= ΔW. W=Fd, and since in both scenarios the same force and same distance is travelled. K1=K2. I am assuming that the objects are at non zero height so by P=mgh, P1>P2

b. Again I am assuming that the objects are at non zero height so by P=mgh, P1>P2.  A heavier mass, a constant force means a smaller acceleration. So a1<a2. We can then use x-x0=v0t+1/2at² and since v0=0, x-x0(d)=1/2at². Solve for t²=2d/a. Since t is the same for both but a1<a2, d1<d2. And since Kinetic Energy=ΔW, W=Fd and F is constant while d1<d2, K1<K2.

The relation will be:(a) K1 = K2(b) K1 < K2

According to the question,

Potential energy be "P".Kinetic energy be "K".

(a)

Word done towards both the block will be similar.

So,

→ [tex]P1 = P2[/tex]

→ [tex]K1= K2[/tex]

(b)

We know,

→ [tex]a = \frac{F}{M}[/tex]

or,

→ [tex]V = a\times t[/tex]

Now,

→ [tex]K = \frac{1}{2} MV^2[/tex]

       [tex]= 0.5\times M\times V^2[/tex]

       [tex]=0.5\times M\times (\frac{F^2}{M^2} )\times t^2[/tex]

       [tex]= 0.5\times F^2\times \frac{t^2}{M}[/tex]

The force and t will be same. So K of the smaller mass will be greater than the larger mass.

hence,

→ [tex]K1<K2[/tex]

Thus the above responses are correct.        

Learn more about friction here:

https://brainly.com/question/13340887

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