The elastic limit of an alloy is 5.0×108 N/m2. What is the minimum radius rmin of a 4.0 m long wire made from the alloy if a single strand is designed to support a commercial sign that has a weight of 8000 N and hangs from a fixed point? To stay within safety codes, the wire cannot stretch more than 5.0 cm.

Explanation:

According to Thermal Expansion of solids:

[tex]dl = \alpha \times l \times dt[/tex]

[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]

[tex]dl = 7.2 \times {10}^{ - 3} [/tex]

Answer:

Explanation:

At the point midway between wires

magnetic field due to wire having current 2I₀

= 10⁻⁷ x 2 x2I₀ / r     where 2r is the distance between wires .

magnetic field due to wire having current I₀

= 10⁻⁷ x 4 I₀ / r

magnetic field due to wire having current I₀

= 10⁻⁷ x 2I₀ / r    

= 10⁻⁷ x 2 I₀ / r     where 2r is the distance between wires .

these fields are in opposite direction as direction of current is same in both .

net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r

= 2 x 10⁻⁷ x  I₀ / r

At point A net magnetic field = 2 x 10⁻⁷ x  I₀ / r

At point B , we shall calculate magnetic field

magnetic field due to nearer wire having current  2 I₀ = 10⁻⁷ x 4 I₀ / r

magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r

These magnetic fields act in the same direction so they will add up

net magnetic field = [ (4 I₀ / r)  + (2 I₀ / 3r) ] x 10⁻⁷

= (14 I₀ / 3r ) x 10⁻⁷

Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷

Ratio of field at A and B

= 3 / 7 . Ans

The ratio of the magnitude of the magnetic field at point A to point B is :

3 / 7

Given data :

Magnitude of the left current is  2I₀

Magnitude of the right current is  I₀

First step : Determine the magnetic field at point A  

The magnetic field due to the left current ( 2I₀ )

10⁻⁷ * 2 * 2I₀ / r       ( 2r = distance between wires )

The magnetic field due to the right current ( I₀ )

10⁻⁷ * 2 I₀ / r

From the expressions above the magnetic fields are in  opposite direction

∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r =   2 * 10⁻⁷ *  I₀ / r

Hence The magnetic field at point A = 2 * 10⁻⁷ *  I₀ / r

Next step : determine the magnetic field at point B

Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r

Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r

Since the fields acts in the same directions

The net magnetic field =  (4 I₀ / r)  + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷

Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷

Therefore the ratio of the magnitude of the magnetic field at point A to point B  =  3/ 7

Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B  = 3 / 7

Learn more : https://brainly.com/question/22403676

Answer:

The bumper will be able to move by 0.01155m.

Explanation:

The magnitude of deceleration of the car in the front end collision.

[tex]a = \frac{F_m}{m} \\[/tex]

[tex]a = \frac{80000}{1500} \\[/tex]

[tex]a = 53.33[/tex]

This is the deceleration of the car that is generated to stop due to a front end collision.

4 km/h = 1.11 m/s

Now, the initial speed of the bumper in the relation of car, Vi = 0

Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s

Use the below equation:

[tex]s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\[/tex]

[tex]s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\[/tex]

[tex]s = 0.01155 \\[/tex]

Thus, the bumper can move relative to the car is 0.01155 m .

Answer:

2,66

Explanation:

The refractive index= real depth/ apparent depth

real depth = refractive index * apparent depth

Let's assume index for water is 1.33

real depth = 2*1,33 = 2,66

Answer:

(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV

(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Explanation:

Given;

radius of the circular loop, r = 31.0 cm = 0.31 m

initial magnetic field, B₁ = 0.7 T

final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T

duration of change in the field, t = 29

(a) The magnitude of induced emf in the loop while the magnetic field is increasing.

[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]

[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]

Where;

A is the area of the circular loop

A = πr²

A = π(0.31)² = 0.302 m²

[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]

(b) the magnitude of the induced voltage at a constant magnetic field

E = A x B/t

E = (0.302 x 1.61) / 3.9

E = 0.1247 V

E = 124.7 mV

Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Answer:

A) y = 3.56 mm

B) y = 3.56 mm

Explanation:

A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:

[tex] y = \frac{m\lambda L}{d} [/tex]

Where:

λ: is the wavelength = 546.1 nm

m: is first bright region = 1

L: is the distance between the screen and the plane of the parallel slits = 1.50 m

d: is the separation between the slits = 0.230 mm

[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]  

B) The distance between the first and second dark bands is:

[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]

Where:

[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]

[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]      

I hope it helps you!

Answer:

Explanation:

The fundamental frequency f₀ is expressed as f₀ =V/2L where;

V is the speed of the string = [tex]\sqrt{\frac{T}{M} }[/tex]

m is the mass of the string

L is the length of the string

T is the tension in the string

f₀ = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

Given datas

m = 5.69g = 0.00569 kg

T = 440N

L = 0.630 m

Required

Fundamental frequency of the steel piano wire f₀

[tex]f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz[/tex]

Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz

Answer:

A) the frequency and amplitude of the output voltag

Explanation:

Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.

Changing the speed regulator will change the engine throttle setting to maintain the speed.

While the power, torque, current, fuel flow rate and torque angle will have decreased.

i

CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

.

a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

Answer:

 f = - 20 cm

Explanation:

This exercise asks us for the focal length, which for a lens in air is

                  1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)

where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.

In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression

                 - 1/40 = (n-1) (1 / R₁ -1 / R₂)

                (1 / R₁ - 1 / R₂) = - 1/40 (n-1)

let's calculate

               (1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)

               (1 / R₁ -1 / R₂) = -1/20

 Now we change the construction material for one with refractive index

n = 2, keeping the radii,

              1 / f = (n-1) (1 / R₁-1 / R₂)

              1 / f = (n-1) (-1/20)

let's calculate

             1 / f = (2.00-1) (-1/20)

              1 / f = -1/20

              f = - 20 cm

Answer:

Answer:

A. Increasing the number of lines per length.

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

Answer:

 λ = 605.80 nm

Explanation:

These double-slit experiments the equation for constructive interference is

          d sin θ = m λ

where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.

In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm

Let's use trigonometry to find the angle

         tan θ = y / L

as the angles are very small

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

we substitute in the first equation

         d y / L = m λ          

          λ = d y / m L

let's calculate

           λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)

           λ = 6.05805 10⁻⁷ m

let's reduce to nm

          λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)

          λ = 605.80 nm

Answer:

The wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm

d. 198 mm

Explanation:

Refractive index is given by;

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}}[/tex]

where;

[tex]\lambda_{vacuum}[/tex] is the wavelength of the light beam in vacuum

[tex]\lambda_{medium}[/tex] is the wavelength of the beam in a material

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}} \\\\\lambda_{vacuum} = \mu *\lambda _{medium}\\\\\ the \ wavelength \ of \ the \ light \ beam \ is \ constant \ in \ a \ vacuum\\\\ \mu_1 *\lambda _{medium}_1 = \mu_2 *\lambda _{medium}_2\\\\\lambda _{medium}_2 = \frac{ \mu_1 *\lambda _{medium}_1 }{ \mu_2} \\\\\lambda _{medium}_2 =\frac{1.5*330}{2.5} \\\\\lambda _{medium}_2 = 198 \ mm[/tex]

Therefore, the wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm.

d. 198 mm

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

             = 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

Answer:

[tex]density \: = \frac{mass}{volume} [/tex]

1 / 5.587 is equal to 0.179 kg/m³

Hope it helps:)

Answer:

The answer is

Explanation:

Density of a substance is given by

[tex]Density \: = \frac{mass}{volume} [/tex]

From the

mass = 1 kg

volume = 5.583 m³

Substitute the values into the above formula

We have

[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]

We have the final answer as

Hope this helps you

Answer:

this raise the temperature is x = 3

Explanation:

Heat capacity is the relationship between heat and temperature change

          C = Q / ΔT

if the heat in the system is given by the change in energy and we carry this differential formulas

          [tex]c_{v}[/tex] = dE / dT

In this problem we are told that the energy of thermal radiation is

        E = A V T⁴

Let's look for the specific heat

        c_{v} = AV 4 T³

the power to which this raise the temperature is x = 3

Answer:

A)  The distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B)  The electrical force of repulsion among them shows a small effect  in beam stability.

Explanation:

Given that:

A Van de Graaff generator produces a beam of 2.02-MeV deuterons

If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.

[tex]\mathtt{K.E = \dfrac{1}{2}mv^2}[/tex]

2 K.E = mv²

[tex]\mathtt{v^2 = \dfrac{2 K.E }{m}}[/tex]

[tex]\mathtt{v =\sqrt{ \dfrac{2 K.E }{m}}}[/tex]

so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.

[tex]\mathtt{v =\sqrt{ \dfrac{2 (2.02 \ MeV) \times \dfrac{10^6 \ eV}{ 1 \ MeV} \times \dfrac{1.60 \times 10^{-19} \ J }{1 \ eV} }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]

[tex]\mathtt{v =\sqrt{ \dfrac{6.464 \times 10^{-13} \ J }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]

v = 13911611.49  m/s

v = 1.39 × 10⁷ m/s

So, If the beam current is 10.0 μA.

We all know that:

[tex]I = \dfrac{q}{t}[/tex]

[tex]t = \dfrac{q}{I}[/tex]

[tex]\mathtt{ t = \dfrac{1.6 * 10 ^{-19} \ C}{10.0 * 10^{-6} \ A}}[/tex]

t = 1.6 × 10⁻¹⁴ s

Finally, the distance of the deuterons from one another  = v × t

the distance of the deuterons from one another  = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)

the distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.

The electrical force of repulsion among them shows a small effect  in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.

Answer:

65.3

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

Answer:

It will be 》》》》1.664716m

Answer:

The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Explanation:

Given;

energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J

speed of light, c = 3 x 10⁸ m/s

Planck's constant, h = 6.63 x 10⁻³⁴ J.s

The wavelength of the emitted light will be calculated by applying energy of photons;

[tex]E = hf[/tex]

where;

E is the energy emitted light

h is Planck's constant

f is frequency of the emitted photon

But f = c / λ

where;

λ is the wavelength of the emitted photons

[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]

λ ≅ 532 nm

the wavelength of the emitted photons is 532 nm.

Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Answer:

0.002699 m or 2.699 mm

Explanation:

y = Fringe distance

d= Distance between slits = 0.310mm

L = Screen distance = 4.40m

λ= Wavelength

Given from question

λ₁= 660 nm = 6.6 x 10^-9 m

λ₂= 470 nm = 4.7 x 10^-9 m

d = 0.340 mm = 3.4 x 10^-3 m

L = 4.40 m

In the case of constructive interference, we use below formula

y/L = mλ/d

For first order wavelength

(y₁/4.40) =(1×660x10⁻⁹)/(0.310*10⁻³)

y₁= (0.310*10⁻³)×(4.40)/(0.310*10⁻³)

y₁=0.00937m

(y2/4.40) =(1×470x10⁻⁹)/(0.310*10⁻³)

y2= =(1×470x10⁻⁹)×(4.40)/(0.310*10⁻³)

y2=0.00667m

distance between the fringes is given by (y₁ -y2)

=0.00937-0.00667=0.002699m

Therefore, distance on the screen between the first-order bright fringes for the two wavelengths is 0.002699 m or 2.699 mm

Answer:

Explanation:

A) 131 dB = 10*log(I / 1e-12W/m²)

where I is the intensity at 2.6 m away.

13.1 = log(I / 1e-12W/m²

1.25e13= I / 1e-12W/m²

I = 1.25 x10^1W/m²

power = intensity * area

P = I * A = 12.5W/m² * 4π(2.6m)² =1061 W ◄

B) 86 dB = 10*log(I / 1e-12W/m²)

8.6 = log(I / 1e-12W/m²)

3.98e8 = I / 1e-12W/m²

I = 3.98e-4 W/m²

area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²

A = 4πr²

2.66e6 m² = 4πr²

r = 14.5m ◄

Answer:

attached below is the free body diagram of the missing  illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

[tex]E_{e} = E_{3} - E_{1}[/tex]

E[tex]_{e}[/tex] = initial kinetic energy of the electron

E[tex]_{1}[/tex] = -4 eV

E[tex]_{3}[/tex] = -1 eV

insert the values into the equation above

[tex]E_{e}[/tex] = -1 -(-4)  eV

   = -1 + 4 = 3 eV

Answer:

South and West

Explanation:

Those people are pushing the hardest. It will move south faster than it moves west.

Answer:

4 x 10¹⁵

Explanation:

Answer:

4 / 7

Explanation:

1/total resistance = 1/1 + 1/2 + 1/4

= 1¾

total resistance = 1 ÷ 1¾

= 4/7

Answer:

The room with the lower temperature

Explanation:

Using

PV=nRT

Since both the rooms same volume and are connected, so they will have same pressure

PV=nRT=constant

nT=Constant/R=constant

If T is more n has to be less

Thus, lower the temperature, more the number molecules.