Two circular, concentric and coplanar turns of
radii R1 = 30 cm and R2 = 20 cm are traversed by currents
i1 = 5 A and i2 = 2 A, indicated in the figure. being the constant of
magnetic permeability in vacuum μ0 = 4^.10-7 Tm/A,
characterize the magnetic induction vector originating at center O.

Two Circular, Concentric And Coplanar Turns Ofradii R1 = 30 Cm And R2 = 20 Cm Are Traversed By Currentsi1

Answers

Answer 1

Answer:

(check the pic)

Explanation:

hope it helps

Two Circular, Concentric And Coplanar Turns Ofradii R1 = 30 Cm And R2 = 20 Cm Are Traversed By Currentsi1

Related Questions

Suppose you walk 13.0 m straight west and then 25.0 m straight south. How far are you from your starting point (in m)

Answers

Answer:

28.2 m

Explanation:

Applying,

Pythagoras theorem,

a² = b²+c²............... Equation 1

Where a = The distance from my starting point to my current point, b = distance walked due west, c = distance walked due south

From the question,

Given: b = 13 m, c = 25 m

Substitute these values into equation 1

a² = 13²+25²

a² = 169+625

a² = 794

a = √794

a = 28.18 m

a ≈ 28.2 m

What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs

Answers

Organs are made up of different types of organs.

Which conclusion can be made based on the information in the table?
Wave speed and wavelengths can vary inversely to produce the same frequency.
O Frequency and wave speed can vary directly to produce the same wavelength.
O Wavelengths and frequency can vary inversely to produce the same wave speed.
O Frequency and wavelengths can vary directly to produce the same wave speed.
Mark this and return
Save and Exit
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Previous Activity

Answers

Answer:

The correct option is (b).

Explanation:

The relation between the wavelength and frequency is given by :

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

Where

v is the wave speed

f is the frequency of a wave

It is clear from the above equation that the wavelengths and frequency can vary inversely to produce the same wave speed.

A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices

Answers

Answer:

Points away from the disk along the z-axis.

Explanation:

Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.

Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.

what units of measurement measures both velocity and speed

Answers

Answer:

[tex]metre \: per \: second[/tex]

Explanation:

Velocity is a derived quantity and the S.I unit is metre per second.Speed is also a derived quantity which is has the S.I unit to be metre per second.

1 point
Q.29. A stone has a weight of 5.7 N.
The gravitational field strength g is 10
N/kg.What is the mass of the stone?
O A 0.57 kg​

Answers

Answer:

weight/mass = gravitational field strength

Given :

Weight of stone = 5.7 N

Gravitational field strength (g) = 10 N/kg

Taking Mass of stone x

=> 5.7/x = 10

x = 10 * 5.7

x = 57 kg

Therefore mass of stone is 57 kg

A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)

Answers

Answer:

The force between the 10 th car and the 11 th car is 13636.4 N.

Explanation:

Force, F = 150 kN

acceleration, a = 2 m/s^2

Let the mass of each car is m. \Total numbers of cars = 11

F = n m a

150000 = 11 x m x 2

m = 6818.18 kg

The force between the 10 th and 11 th car is

T = ma = 6818.18 x 2 = 13636.4 N

which team won the champions league in 2020 2021​

Answers

Answer:

Chelsea F.C

Explanation:

Chelsea F.C

Soccer

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity.

Required:
a. A plane accelerates from rest at a constant rate of 5.00m/s^2 along a runway that is 1800m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off?
b. What is the distance dfirst traveled by the plane in the first second of its run?
c. What is the distance dfirst traveled by the plane in the first second of its run?

Answers

Answer:

(a)

67.1 s

(b) 2.5 m

(c) 2.5 m

Explanation:

initial speed, u = 0 m/s

final speed, v = 70 m/s

acceleration, a = 5 m/s2

distance, s = 1800 m

(a) Use third equation of motion

[tex]v^2= u^2 + 2 a s \\\\v^2 = 0 + 2 \times 5\times 1800\\\\v =134.2 m/s[/tex]

Let the time is t.

Use first equation of motion

v = u + at

134.2 = 0 + 5 t

t = 67.1 s

(b) Use second equation of motion

[tex]s = u t +0.5 at^2\\\\s = 0 +0.5\times 5\times 1 \\\\s = 2.5 m[/tex]

(c) Use second equation of motion

[tex]s = u t +0.5 at^2\\\\s = 0 +0.5\times 5\times 1 \\\\s = 2.5 m[/tex]

Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by Icm= 1/2mL^2.

Required:
Find Icmd the moment of inertia of the rod with respect to a parallel axis through one end of the rod.

Answers

Answer:

right now I see some of you have a great day

Answer: (mI^2)/3

Explanation:

The parallel axis theorem for the calculation of inertia is: I = I CM + Md^2

So, I is the apathy from an axis that is at distance d from the center of mass and LCM the apathy when the axis passes through the center of mass. Do to this, the axis passes through the end of the rod. In analysis, d=l/2

So, we have the equation:

I = mI^2/12 + m (1/2)^2 = mI^2/12 + mI^2/4 = mI^2/12 + 3mI^2/12 = mI^2/3

This relents us the terminal result: (ml^2)/3.

A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .

Required:
How far from the second lens is the final image of an object infinitely far from the first lens?

Answers

Answer:

the required distance is 6 cm

Explanation:

Given the data in the question;

f₁ = 15 cm

f₂ = 5.0 cm

d = 45 cm

Now, for first lens object distance s = ∝

1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'

Now, image distance of first lens s' = 15cm  

object distance of second lens s₂ will be;

s₂ = 45 - 15 = 30 cm

so

1/f₂ = 1/s₂ + 1/s'₂

1/5 = 1/30 + 1/s'₂

1/s'₂ = 1/5 - 1/30  

1/s'₂ = 1 / 6

s'₂ = 6 cm

Hence, the required distance is 6 cm

 

The distance of the final image from the first lens will be is 6 cm.

What is mirror equation?

The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).

The given data in the problem is;

f₁ is the focal length of lens 1= 15 cm

f₂ s the focal length of lens 2= 5.0 cm

d is the distance between the lenses = 45 cm

From the mirror equation;

[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]

If f₁ is the focal length of lens 1 is 15 cm then;

[tex]s'=15 cm[/tex]

f₂ s the focal length of lens 2= 5.0 cm

s₂ = 45 - 15 = 30 cm

From the mirror equation;

[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]

Hence the distance of the final image from the first lens will be is 6 cm.

To learn more about the mirror equation refer to the link;

https://brainly.com/question/3229491

Celestial Events, such as rise, set or transit times are represented by the intersection of various diagonal lines (and loops) with the horizontal and vertical lines, this will allow us to determine what about the Celestial Event?
a) Distance
b) Latitude
c) Time and Date
d) Gamma Rays

Answers

Answer:

C) time and date

Explanation:

Celestial event is an astronomical phenomenon. This involves the conjunction of one or more celestial objects such as lunar and solar eclipse or meteor shower. The intersecting horizontal and vertical lines allow the astrologists to determine the time and date of the celestial event.

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time

Answers

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *

Answers

Answer:

you drove 50km

Explanation:

10×5 hope this helps

Answer:

50 Km

Explanation:

This is how far you have got on your journey if traveling like this.

Please Mark as Brainliest

Hope this Helps

Please help I need this done within 30 mins

Answers

It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.

A jogger moves from a position x =
4.0 m to a position of x = 16.0 m in
4.0 s. What was her average velocity?
(Unit = m/s)
Don't forget: velocities and displacements to
the right are +, to the left are -,

Please help me!

Answers

Answer:

3 m/s

Explanation:

We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:

Initial displacement (d₁) = 4 m

Final displacement (d₂) = 16 m

Change in displacement (Δd) =?

Δd = d₂ – d₁

Δd = 16 – 4

Δd = 12 m

Finally, we shall determine the determine the average velocity. This can be obtained as follow:

Change in displacement (Δd) = 12 m

Time (t) = 4 s

Velocity (v) =?

v = Δd / t

v = 12 / 4

v = 3 m/s

Thus, the average velocity of the jogger is 3 m/s

7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
driver of the car applies the brakes after a reaction time of 0.5 s and the car slows down at a rate of
10 ms. What happens?

Answers

Answer: Car collide with man

Explanation:

Given

Speed of car is [tex]u=30\ m/s[/tex]

Distance of the man from the car is [tex]s=55\ m[/tex]

Reaction time [tex]t_r=0.5\ s[/tex]

Rate of deceleration [tex]a_d=-10\ m/s^2[/tex]

Distance traveled in the reaction time [tex]d_o=30\times 0.5=15\ m[/tex]

Net effective distance to cover [tex]d=55-15=40\ m[/tex]

Distance required to stop the car

[tex]\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m[/tex]

Require distance is more than that of net effective distance. Hence, car collides with the man.

How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *

Answers

Answer:

50km

Explanation:

So we know we drove 5km a hour.

We drove like this for 10 hours.

To find the total km, we must multiply the time driven by the speed:

time*speed=distance

Plug in our values:

10*5km=50km

So your answer is 50km.

Hope this helps!

The answer is 50km/hr

If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?

Answers

Answer:

72  is the premimum of the insurance.

Explanation:

Below is the given values:

The loading = 0.4

Coinsurance rate = 0.2

Number of units = 100

Total number of units = 100 * 0.4 = 40

Remaining units = 60 * 0.2 = 12

Add the 60 and 12 values = 60 + 12 = 72

Thus, 72  is the premimum of the insurance.

g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.

Answers

Answer:

[tex]V=8.08m/s[/tex]

Explanation:

From the question we are told that:

Height[tex]h=5.00m[/tex]

Mass [tex]m=0.750kg[/tex]

Radius [tex]r=4.00cm=>0.04m[/tex]

Generally the equation for Total energy is mathematically given by

  [tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]

Therefore

 [tex]V=\sqrt{\frac{4gh}{3}}[/tex]

 [tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]

 [tex]V=8.08m/s[/tex]

Define Potential Energy
Begin by defining potential energy in your own words within one concise eight word sentence

Answers

Answer:

potential energy is a type of energy an object has because of it's position

A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction

Answers

Answer:

The kinetic energy and potential energy lost to friction is 2,420 J.

Explanation:

Given;

total mass, m = 40 kg

initial velocity of the girl, Vi = 5 m/s

hight of the hill, h = 10 m

length of the hill, L = 100 m

initial kinetic energy of the girl at the top hill:

[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]

initial potential energy of the girl at the top hill:

[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]

Total energy at the top of the hill:

E = 500 J + 3920 J

E = 4,420 J

At the bottom of the hill:

final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s

hight of the hill = 0

final kinetic energy of the girl at the bottom of the hill:

[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]

final potential energy of the girl at the bottom of the hill:

[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]

Based on the principle of conservation of energy;  

the sum of the energy at the top hill = sum of the energy at the bottom hill

The energy at the bottom hill is less due to energy lost to friction.

[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]

Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.

A cat pushes a porcelain statue off a bookshelf with a speed of 0.5 m/s and it smashed on the floor 0.85 sec later.

Answers

Answer:

167?

Explanation:

i added both

16. The sum of kinetic energies in an object.
17. The essential device in power plants that convert mechanical
energy to electricity.
18. The device that converts electricity back to mechanical energy
19. The only EM wave that is seen by naked eye.
20. A device that converts light to electricity.​

Answers

Yes I also need help on this

plz answer fast the question

Answers

Answer:

Angle of incidence = 20°

Angle of reflection = 20°

Explanation:

Applying,

The first Law of Refraction: The incident ray, the reflected ray and the normal at the point of incidence all lies in the plane.

From the diagram,

Angle of incidence = 90-70

Angle of incidence = 20°

From the law of reflection,

Angle of incidence = Angle of reflection

Therefore,

Angle of reflection = 20°

True or False. A person who is nearsighted cannot see objects that are close to them clearly.

Answers

false, farsightedness is when you cant see close

Answer:

false

Explanation:

hope it works

You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 1.8 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 28.0 m/s when they reach the bottom of the ramp. You determine that for a 75kg skier with good form, friction and air resistance will do total work of magnitude 3500 J on him during his run down the slope. What is the maximum height (h) for which the maximum safe speed will not be exceeded?

Answers

Answer:

44.6 m

Explanation:

From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.

E = E'

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J

Substituting the values of the variables into the equation, we have

U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂

mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J

9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J

(735 kgm/s²)h + 75  kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J

(735 kgm/s²)h + 121.5  kgm²/s² = 29400 kgm²/s² + 3500 J

(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J

(735 kgm/s²)h + 121.5 J = 32900 J

(735 kgm/s²)h = 32900 J - 121.5 J

(735 kgm/s²)h = 32778.5 J

h = 32778.5 J/735 kgm/s²

h = 44.6 m

So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

Answers

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Explanation:

It is known that formula for path difference is as follows.

[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex]    ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.

[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]

Hence, path difference is as follows.

[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]

For lowest frequency, the value of n = 0.

[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]

[tex]\lambda = 4 \Delta L[/tex]

where,

[tex]\lambda[/tex] = wavelength

The relation between wavelength, speed and frequency is as follows.

[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]

where,

[tex]\nu[/tex] = speed

f = frequency

Substitute the values into above formula as follows.

[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

I need help with physics question.

Answers

(D)

Explanation:

Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is

F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)

= 2.0×10^-11 N

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Force and power. Part A How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hphp

Answers

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm  

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

[tex]P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580[/tex]

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