A stationary horn emits a sound with a frequency of 228 Hz. A car is moving toward the horn on a straight road with constant speed. If the driver of the car hears the horn at a frequency of 246 Hz, then what is the speed of the car? Use 340 m/s for the speed of the sound

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

[tex]R = \frac{\rho L}{A}[/tex]

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]

L = 169.5 m

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

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

k = mg/x  

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

[tex]\phi = BAcos \theta[/tex]

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]

Hence the magnetic flux Φ through the loop is 0.5849Weber

Answer:

The correct answer is - 8.99N/C

Explanation:

[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]

Answer:

Magnification = 1

Explanation:

given data

radius of curvature r = - 0.983 m

image distance u = - 0.155

solution

we get here first focal length that is

Focal length, f = R/2     ...................1

f = -0.4915 m

we use here formula that is

[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex]      .................2

put here value and we get

[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]  

so

Magnification will be here as

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

m =  [tex]\frac{0.155}{0.155}[/tex]

Answer:

The magnification is 1.5.

Explanation:

radius of curvature, R = - 0.983 m

distance of object, u = - 0.155 m

Let the distance of image is v.

focal length, f = R/2 = - 0.492 m

Use the mirror equation

[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]

The magnification is given by

m = - v/u

m = 0.226/0.155

m = 1.5

Answer:

So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.

[tex]F = 5.93×10^{13}\:\text{N}[/tex]

Explanation:

Given:

[tex]m_1= 2×10^{16}\:\text{kg}[/tex]

[tex]m_2= 4×10^{22}\:\text{kg}[/tex]

[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]

Using Newton's universal law of gravitation, we can write

[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]

[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]

[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

Answer:

d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.

Explanation:

That is a problem of electric forces, given by Coulomb's law

          F = [tex]k \frac{ q1q2}{r^2}[/tex]

We use that charges of the same sign repel and charges of different signs do not attract, so the net force is

           ∑ = F₁₃ + F₂₃

          F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]

a) the charge is placed at the midpoint between the other two

         F_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]

calculate us

          F_ {net} = 9 10⁹ / 0.5²   2 10⁻⁶ (50 -25) 10⁻⁶

          F_ {net} = 1,800 N

b) where must be placed q3 so that the force is zero

for this case the charge q3 is outside the spheres

          ∑ F = 0

          F₁₁₃ = F₂₃

          k q_1 / r_{13}² = k q₂ q₃ / r₂₃²

          q₁/ r₁₂²   = q₂ / r₂₃²

suppose the distance

          r₁₂ = d

the he other sphere is

          r₂₃ = d + 1

we substitute

          q₃ / d² = q₂ / (d + 1) ²

          (d + 1) ² = q₂ / q₃ d²

           d² (1 - q₂/ q₃) + 2d + 1 = 0

we solve the equation of a second

            d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2

             d = -2 /2

             d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.

Answer:

elevators

Theatre system

construction pulley

lifts

Answer:

elevator,cargo lift system

Answer:

The percentage of people who wanted both the swimming pool and the soccer complex is 0.6 + 0.6 – 0.95 = 0.25. This can also be seen in the Venn diagram.

Explanation:

Edmentum

The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ[tex]T^{4}[/tex]

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.

So that;

[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]

     = 3.8556 x [tex]10^{-8}[/tex]

     = 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex]

Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

Learn more on energy radiation of objects by visiting: https://brainly.com/question/12550129

Being in parallel each device will have an equal voltage drop of 120 V

A. Yes the combination will blow the fuse. See part B for the total current.

B. Toaster = 1800W / 120V = 15A

   Frying Pan = 1400W / 120V = 11.67A

  Lamp = 55W / 120V = 0.458A

Total amps = 15 + 11.67 + 0.458 = 27.128 Amps

27.128A is greater than 15A so the fuse will blow.

Answer:

to adjust from distant to the near objects

Explanation:

Answer:

The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.

Explanation:

The human eye is the optical instrument which works on the refraction of light.

The ability of eye lens to change its focal length is called accommodation power of eye.

The focal length of eye lens is changed by the action of ciliary muscles.

When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.

Answer:

The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.

Explanation:

285.3 days

Explanation:

The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write

[tex]F_c = F_G[/tex]

or

[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]

where M is the mass of the star and R is the orbital radius around the star. We know that

[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]

where C is the orbital circumference and T is orbital period. We can then write

[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]

Isolating [tex]T^2[/tex], we get

[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]

Taking the square root of the expression above, we get

[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]

which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as

[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]

[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

Answer:

[tex]N_f=248N[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=100kg[/tex]

Ladder Length [tex]l=4.0m[/tex]

Mass of Ladder [tex]m_l=25kg[/tex]

Angle [tex]\theta=56 \textdegree[/tex]

Generally the equation for Co planar forces is mathematically given by

[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]

Therefore

[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]

[tex]N_f=248N[/tex]

Answer:

N/C = V/m.

Explanation:

The SI unit of electric field is N/C. Sometimes the units are written as V/m.

We know that,

1 V = 1 J/C

Using dimensional analysis,

The dimensional formula for Joules is [M¹L² T⁻²].

The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].

So,

J/C = [M¹L² T⁻³I¹] ...(1)

The dimensional formula of Newton is [M¹ L¹ T⁻²]

The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].

N/C= [M¹L² T⁻³I¹] ....(2)

From (1) and (2) it is clear that the N/C is equal to V/m.

Answer:

International System of Units. it was established in 1960 by the 11the General Conference on Weights and Measures

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

         p₀ = m v₁₀ + 0

         p₀ = 0.6 2

         p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

        p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

         p₀ₓ = 1.2 kg m / s

         p_{fx} = m v1f cos 20 + m v2f cos θ

         p₀ = p_f

        1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

         1.2482 = v_{2f} cos θ

Y axis  

        p_{oy} = 0

        p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

        0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

        0.2736 = v_{2f} sin θ

we write our system of equations

         0.2736 = v_{2f} sin θ

         1.2482 = v_{2f} cos θ

divide to solve

         0.219 = tan θ

          θ = tan⁻¹ 0.21919

          θ = 12.36

let's look for speed

            0.2736 = v_{2f} sin θ

             v_{2f} = 0.2736 / sin 12.36

            v_{2f} = 1.278 m / s

Answer:

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Explanation:

Answer:

The current in third branch is 11 A.

Explanation:

incoming current in one branch = 18 A

outgoing current in the other branch = 7 A

let the current in the third branch is i.

According to the Kirchoff's fist law in electricity

incoming current = out going current

18 = 7 + i

i = 11 A

The current in third branch is 11 A.

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.

Answer:

Increasing until terminal velocity is reached

Explanation:

Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.

Answer:

Explanation:

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Answer:

Can't understand the language