name a device that converts mechanical energy into electrical energy.

Since the frequencies are unrelated, and there are a large number of them, I'll say this represents an example of noise.

Answer:

This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency

The acceleration of the block over the rough surface is 1.22625 m/s²

The process through which the acceleration is obtained is presented as follows of approach to

The given parameters are;

Mass of block, m = 2 kg

Nature of the surface of the quarter circle = Frictionless

The length of the horizontal, d = 8 m

The coefficient of friction of the horizontal surface, μ = 0.4

The unknown parameter;

The acceleration of the block over the rough surface

Method;

Find the work done by friction to stop the block and divide the result by the mass of the block

The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)

[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d

[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ

Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block

[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N

Therefore;

The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d

[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J

The work done by the block, W = Force, F × d

Force, F = m × a

Where;

a = The acceleration of the block

According to the principle of conservation of energy, we have;

[tex]\mathbf{W_f}[/tex]  = W

∴ 19.62 J = 2 kg × a × 8 m

a = 19.62/(2 kg × 8 m) = 1.22625 m/s²

The acceleration of the block over the rough surface, a = 1.22625 m/s²

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Answer:

15.88

is the correct answer

Answer:

it will drop simultaneously

Answer:

The amplitude of vibration of the spring is "24 cm"

The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.

From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.

Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.

Amplitude = 24 cm

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

Determine the mass transfer coefficient ( m/s )

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

Answer:

Electric field at A = 9.28 x 10¹² N/C

Explanation:

Given:

K = 8.99 x 10¹² N/C

Missing information:

Length = 11 cm = 11 x 10⁻² m

q = 12.5 C

Find:

Electric field at A

Computation:

Electric field = Kq / r²

Electric field at A = [(8.99 x 10¹²)(12.5)] / [11 x 10⁻²]²

Electric field at A = 9.28 x 10¹² N/C

Answer:

rock go down

Explanation:

what comes up must come down.

Answer:

a)  fem = - 2.1514 10⁻⁴ V,  b) I = - 64.0 10⁻³ A, c)    P = 1.38  10⁻⁶ W

Explanation:

This exercise is about Faraday's law

         fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]

where the magnetic flux is

        Ф = B x A

the bold are vectors

        A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

         fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

        v = w r

        w = 2π f

        v = 2π f r

we substitute

        fem = - 2π B r (2π f r)

        fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

       f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

       fem = - 4π² 428 10⁻⁶ 4π 0.10²

       fem = - 16π³ 428 10⁻⁶ 0.010

       fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

        V = I R

        I = V / R

         I = -2.1514 10⁻⁴ / 0.00336

         I = - 64.0 10⁻³ A

Electric power is

        P = V I

        P = 2.1514 10⁻⁴ 64.0 10⁻³

        P = 1.38  10⁻⁶ W

Answer:

375 is the answer.

Explanation:

Speed : Distance / Time taken

S: m/ s

s: 1500/4

375 m / s answer

Answer:

375m per minute

Explanation:

if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame

Answer:

g = 0.0114 m/s²

Explanation:

The value of acceleration due to gravity on the surface of the moon can be given by the following formula:

[tex]g = \frac{Gm}{r^2}[/tex]

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m = mass of moon = 4.2 x 10¹⁶ kg

r = radius of moon = 1.57 x 10⁴ m

Therefore,

[tex]g= \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(4.2\ x\ 10^{16}\ kg)}{(1.57\ x\ 10^4\ m)^2}[/tex]

g = 0.0114 m/s²

Answer:

[tex]\theta=34 \textdegree[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=55kg[/tex]

Angle [tex]\theta =28.0[/tex]

Coefficient of static friction [tex]\alpha =0.680[/tex]

Generally, the equation for Newtons second Law is mathematically given by

For

[tex]\sum_y=0[/tex]

[tex]N=mgcos \theta[/tex]

for

[tex]\sum_x=0[/tex]

[tex]F_{s}=mgsin\theta[/tex]

Where

[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]

[tex]F_{s}=0.68*55*9.8*cos 28[/tex]

[tex]F_{s}=323.62N[/tex]

Therefore

[tex]\alpha mgcos \theta=mg sin \theta[/tex]

[tex]\theta=tan^{-1}(0.68)[/tex]

[tex]\theta=34 \textdegree[/tex]

(a) The static frictional force which holds the box in place is 323.62 N.

(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.

The net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.

∑F = 0

The static frictional force is calculated as follows;

Fs = μFncosθ

Fs = 0.68 x (55 x 9.8) x cos28

Fs = 323.62 N

Fn(sinθ) - μFn(cosθ) = 0

mg(sinθ) - μmg(cosθ) = 0

μmg(cosθ) = mg(sinθ)

μ(cosθ) = (sinθ)

μ = sinθ/cosθ

μ = tanθ

θ = tan⁻¹(μ)

θ = tan⁻¹(0.68)

θ = 34.2⁰

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Answer:

1.2 seconds

Explanation:

Answer to the following question is 1.2 seconds

Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.

Answer:

[tex]P_d=6203.223062W/mm^2[/tex]

Explanation:

From the question we are told that:

Voltage [tex]V=45kV[/tex]

Current [tex]I=50mAmp[/tex]

Diameter  [tex]d=0.50mm[/tex]

Heat transfer factor [tex]\mu= 0.87.[/tex]

Generally the equation for  Power developed is mathematically given by

[tex]P=VI\\\\P=45*10^3*50*10^{-3}[/tex]

[tex]P=2.250[/tex]

Therefore

Power in area

[tex]P_a=1400*0.87[/tex]

[tex]P_a=1218watt[/tex]

Power Density

[tex]P_d=\frac{P_a}{Area}[/tex]

[tex]P_d=\frac{1218}{\pi(0.5^2/4)}[/tex]

[tex]P_d=6203.223062W/mm^2[/tex]

Explanation:

Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X

The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.

Given:

The number of lines = 6000 per cm,

The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m

Calculate the diffraction grating,

[tex]d = 1 / no\ of\ lines[/tex]

d = 10⁻² / 6000 m,

Calculate the second-order maxima angle and third-order maxima angle by the formula given below,

[tex]dsin\theta_1 = n_1 \lambda[/tex]

[tex]sin\theta_1 = n_1\lambda / d[/tex]

[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]

θ₁ = sin⁻¹(0.6)

θ₁ = 36.87°

Similarly, for θ₂,

θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)

θ₂ = sin⁻¹(0.9)

θ₂ = 64.16°

Calculate the separation as follows,

θ₂ - θ₁ = 64.16° - 36.87°

θ₂ - θ₁ =  27.29°

Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

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Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

Answer:

The angle is 153.9 degree.

Explanation:

Let the magnetic field is B and the charge is q. Angle = 26.1 degree

The force is F.

Let the angle is A'.

Now equate the magnetic forces

[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

[tex]Q_c = 110 \ J[/tex]

Coefficient of performance refrigerator,

[tex]Cop(refrig)=4[/tex]

(a)

As we know,

⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]

or,

⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]

              [tex]=\frac{110}{4}[/tex]

              [tex]=27.5 \ Joules[/tex]

(b)

⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]

or,

⇒ [tex]Q_h = Q_c+Work[/tex]

         [tex]=114+27.5[/tex]

         [tex]=141.5 \ Joules[/tex]

The resistance R = 20 Ω

The capacitance C = 106.1 μF

The current, I is 2.773 A at 56.31°.

The phase angle of the between the current and the voltage is 56.31° leading.

Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω

So, the resistance R = 20 Ω

To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.

Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance  

So, C = 1/ωX = 1/2πfX

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

C = 1/2πfX

C = 1/(2π × 50 Hz × 30 Ω)

C = 1/3000π

C = 1/9424.778

C = 1.061 × 10⁻⁴ F

C = 106.1 × 10⁻⁶ F

C = 106.1 μF

So, the capacitance is 106.1 μF

The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.

The magnitude of Z = √(20² + (-30)²)

= √(400 + 900)

= √1300

= 36.06 Ω

and its angle Φ = tan⁻¹(ImZ/ReZ)

= tan⁻¹(-30/20)

= tan⁻¹(-1.5) = -56.31°

So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°

So, the current, I = V/Z =  (100 ∠ 0°)/36.06 ∠ -56.31°

= 100/36.06 ∠(0° - (-56.31° ))

= 2.773 ∠ 56.31° A

So, the current is 2.773 A at 56.31°.

Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.

So, the phase angle of the between the current and the voltage is 56.31° leading.

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Answer:

The initial velocity is 1.27 m/s.

Explanation:

distance, s = 1.8 m

acceleration, a = - 0.45 m/s^2

final velocity, v = 0

let the initial velocity is u.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]

We have that the Initial velocity  is mathematically given as

u=1.27m/s

Question Parameters:

a slight push toward an end stop 1.80 meters away

he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2

Generally the equation for the  third equation of motion    is mathematically given as

Vf^2 = Vi^2 + 2ad

Therefore

0=u^2+0.45*1.8

u=1.27m/s

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Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

Answer:

The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Explanation:

From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h

The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,

Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s

First, convert 118 km/h to m/s

118 km/h = (118 × 1000) /3600 = 32.7778 m/s

∴ u = 32.7778 m/s

Now, to determine the deceleration, a, required to stop,

From one of the equations of motion for linear motion,

v² = u² + 2as

Then

0² = (32.7778)² + 2×a×85

0 = 1074.3841 + 170a

∴ 170a = - 1074.3841

a = - 1074.3841 / 170

a = - 6.3199

a ≅ - 6.32 m/s²

Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Answer:

explained

Explanation:

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.

A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.

Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.

Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at  5  m/s

straight toward the goal and drives the ball in the same direction with a velocity of  30 m/s

relative to her body, then the velocity of the ball is  35  m/s

relative to the stationary, profusely sweating goalkeeper standing in front of the goal.

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (

The figure shows components of velocity v in horizontal  vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.

Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors  and  

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn [tex]f_o=228\ Hz[/tex]

Apparent frequency [tex]f'=246\ Hz[/tex]

Speed of sound is [tex]V=340\ m/s[/tex]

Doppler frequency is

[tex]\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)[/tex]

Where,

[tex]v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}[/tex]

Insert values

[tex]\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s[/tex]

Thus, the speed of the car is [tex]26.84\ m/s[/tex]

Explanation:

mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.

Answer:

p = 60.6N*s

Explanation:

v_f = v_0+a*t

a = (v_f-v_0)/t

a = (1.8m/s)/2.35s

a = 0.77m/s²

F = m*a

F = (25kg+8.5kg)*0.77m/s²

F = 25.8N

^p = F*t

p = 25.8N*2.35s

p = 60.6N*s

Answer:

prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)

please sove step by step with language it is opt maths question

Answer:

test her prototype and collect data about its flight