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.

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.

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.

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.

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!

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.

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

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

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.

Answer:

Chelsea F.C

Explanation:

Chelsea F.C

Soccer

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

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.

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°

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

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]

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.

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

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.

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]

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.

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]

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.

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

Answer:

167?

Explanation:

i added both

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.

(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

Answer:

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

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

Answer:

false

Explanation:

hope it works

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.