During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an angular displacement of 1.5758 rev. What is the angular acceleration of the CD

Answer:

x(t) = -3sin2t

Explanation:

Given that

Spring force of, W = 720 N

Extension of the spring, s = 4 m

Attached mass to the spring, m = 45 kg

Velocity of, v = 6 m/s

The proper calculation is attached via the image below.

Final solution is x(t) = -3.sin2t

Answer:

9.53 m

Explanation:

The computation of shortest organ pipe with both ends open that will resonate at this frequency is shown below:-

[tex]\lambda = \frac{velocity}{frequency}[/tex]

[tex]= \frac{343}{18}[/tex]

= 19.06 m

Now the

Shortest organ pipe  with both ends open is

=  [tex]\frac{\lambda}{2}[/tex]

[tex]= \frac{19.06}{2}[/tex]

= 9.53 m

Basically we applied the above formulas so that first we easily determined the shortest organ pipe for both ends at this frequency

Answer:

B

Explanation:

ANSEWER :B IN ROWS ONLY

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

[tex]\theta=90-40=50^o[/tex]

From the triangle OAB

[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]

[tex]1340.57=35^2+25^2-x^2[/tex]

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

Answer:

An increase in frequency will cause an increase in the number of lines per centimeter and a smaller distance between each consecutive line. That is the distance between each trough

Explanation:

This is because higher frequency light source should produce an interference pattern with more lines per centimeter in the pattern and a smaller spacing between lines.

Answer:

a) yes

b) no

c) yes

d)Cart 2 with mass [tex]\frac{M}{3}[/tex]   is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁=  M

u₁ = 16m/s

m₂ =[tex]\frac{M}{3}[/tex]

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)

therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)

∴u₂ = 3×16 = 48 m/s

Answer:

Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.

Explanation:

Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.  

Answer:

107 m down the incline

Explanation:

Given:

v₀ₓ = 25 m/s

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

-Δy/Δx = tan 35°

Find: d

First, find Δy and Δx in terms of t.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) t + ½ (-10 m/s²) t²

Δy = -5t²

Δx = v₀ₓ t + ½ aₓ t²

Δx = (25 m/s) t + ½ (0 m/s²) t²

Δx = 25t

Substitute:

-(-5t²) / (25t) = tan 35°

t/5 = tan 35°

t = 5 tan 35°

t ≈ 3.50 s

Now find Δy and Δx.

Δy ≈ -61.3 m

Δx ≈ 87.5 m

Therefore, the distance down the incline is:

d = √(x² + y²)

d ≈ 107 m

Complete Question:

A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.

(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its

speed is 3.7 m/s, what angle does the cord make with the vertical?

(Check attached image for the diagram.)

Answer:

(a) The ball’s speed, v = 2.06 m/s

(b) The angle the cord makes with the vertical is 50.40⁰

Explanation:

If the ball is revolved in a horizontal plane, it will form a circular trajectory,

the radius of the circle, R = Lsinθ

where;

L is length of the string

The force acting on the ball is given as;

F = mgtanθ

This above is also equal to centripetal force;

[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]

(b) when the speed is 3.7 m/s

[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]

[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]

Therefore, the angle the cord makes with the vertical is 50.40⁰

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]          (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by:

[tex]v_{max}=\omega A=2\pi f A[/tex]     (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by:

[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]

[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is:

[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]

The total energy is 0.1J

(e) The displacement as a function of time is:

[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Answer:

[tex]v_{2}=3.5 m/s[/tex]

Explanation:

Using the conservation of energy we have:

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

Let's solve it for v:

[tex]v=\sqrt{2gh}[/tex]

So the speed at the lowest point is [tex]v=7 m/s[/tex]

Now, using the conservation of momentum we have:

[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

[tex]v_{2}=\frac{1*7}{2}[/tex]

Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]

I hope it helps you!

Answer:

The greater the density of the electric field lines the stronger the electric field and vice versa

Explanation:

Electric field can be defined as the region where an electric force is experienced by a charged body. A charged body experiences a force whenever it is positioned close to another charged body.

An electric field may be described in terms of lines of force which represent the direction of a small positive charge placed at that point assuming that the charge is so small that it does not change appreciably in the presence of another charge. Arrows on the lines of force indicate the direction of the electric field.

The lines of force are indicated in such a way that the strength of the electric field is shown by the number or density of electric field lines crossing a unit area perpendicular to the lines. Hence, the greater the density of the electric field lines the stronger the the electric field and vice versa

Answer:

I believe the answer is in fact section (VW) on the line where the electric field result will be zero.

Explanation:

The direction of the electric field due to a positive charge is away from it and the direction of the electric field due to a negative one is towards it.

Answer:

By using 4 resistors in parallel and followed by a resistor in series.

Explanation:

Let consider 5 resistors with a resistance of 20 ohms each, the total resistance is equal to 25 ohms when the first four resistance are settled in parallel and is followed by a resistor is series. The calculations are presented below:

1st Stage (4 resistors in parallel)

[tex]R_{eq,1} = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2} }+\frac{1}{R_{3}}+\frac{1}{R_{4}}}[/tex]

If [tex]R_{1} = R_{2} = R_{3} = R_{4} = 20\,\Omega[/tex], then:

[tex]R_{eq,1} = \frac{1}{\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega} }[/tex]

[tex]R_{eq,1} = \frac{1}{\frac{4}{20\,\Omega} }[/tex]

[tex]R_{eq,1} = 5\,\Omega[/tex]

2nd Stage (1 resistor in series)

[tex]R_{5} = 20\,\Omega[/tex]

Now, the equivalent resistance is:

[tex]R_{eq,2} = R_{eq,1} + R_{5}[/tex]

[tex]R_{eq,2} = 5\,\Omega + 20\,\Omega[/tex]

[tex]R_{eq,2} = 25\,\Omega[/tex]

Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]

where

[tex]S_A[/tex] indicates the speed of the ship A

[tex]S_B[/tex] indicates the speed of the ship B

and

C indicates the velocity of life

now we will Substitute 0.40c for A and 0.60 for B in the equation which is

[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]

after solving the above equation we will get

0.80 c

So, The correct answer is 0.80c

Answer:

Explanation:

Find the diagram relating to the question for proper explanation of the question below.

Using the principle of moment

Sum of clockwise moments = Sum of anticlockwise moments

Moment = Force * perpendicular distance

For anti-clockwise moment:

Since the 30 kg moves in the anticlockwise direction according to the diagram

ACW moment = 30 * 1 = 30 kgm

For clockwise moment

If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).

Equating both moments we have;

0.75M = 30

M = 30/0.75

M = 40 kg

The second child's mass is 40 kg

Answer:

2.45 m/s

Explanation:

kinetic energy = 1/2 * m * v^2

then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2

by solving for x, X = 2.45 m/s

Answer:

 θ₁ = 0.04º , θ₂ = 0.00118º

Explanation:

The equation that describes the diffraction pattern of a network is

             d sin θ = m λ

where the diffraction order is, in this case they indicate that the order

m = 1

           θ = sin⁻¹ (λ / d)

Trfuvsmod ls inrsd fr ll red s SI units

           d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm

the distance between two lines we can look for it with a direct proportions rule

If there are 4724 lines in a centimeter, the distance for two hundred is

            d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm

let's calculate the angles

λ = 300 10-9 m

            θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)

            θ₁ = sin⁻¹ (7.08 10-4)

            θ₁ = 0.04º

λ = 5000

          θ₂ = sin-1 (500 10-9 / 4,2337 10-4)

          θ₂ = 0.00118º

Answer:

109.32 N/m

Explanation:

Given that

Mass of the hung object, m = 8 kg

Period of oscillation of object, T = 1.7 s

Force constant, k = ?

Recall that the period of oscillation of a Simple Harmonic Motion is given as

T = 2π √(m/k), where

T = period of oscillation

m = mass of object and

k = force constant if the spring

Since we are looking for the force constant, if we make "k" the subject of the formula, we have

k = 4π²m / T², now we go ahead to substitute our given values from the question

k = (4 * π² * 8) / 1.7²

k = 315.91 / 2.89

k = 109.32 N/m

Therefore, the force constant of the spring is 109.32 N/m

Explanation:

(a) The acceleration is 9.8 m/s² downwards.

(b) At the maximum height, the velocity is 0 m/s.

(c) v = at + v₀

0 m/s = (-9.8 m/s²) (1.91 s) + v₀

v₀ = 18.7 m/s

(d) Δy = vt − ½ at²

Δy = (0 m/s) (1.91 s) − ½ (-9.8 m/s²) (1.91 s)²

Δy = 17.9 m

Answer:

[tex]V_{f}[/tex] = 6 cm/s

Explanation:

Given:

Acceleration = a = 2 cm/s²

Time = t = 3s

Initial Velocity = [tex]V_{i}[/tex] = 0 cm/s

Required:

Velocity = [tex]V_{f}[/tex] = ?

Formula:

a = [tex]\frac{V_{f}-V_{i}}{t}[/tex]

Solution:

2 = [tex]\frac{V_{f}-0}{3}[/tex]

=> [tex]V_{f}[/tex] = 2*3

=> [tex]V_{f}[/tex] = 6 cm/s

Answer:

τ = 0.26 N.m

Explanation:

First we find the moment of inertia of the wheel, by using the following formula:

I= mr²

where,

I = Moment of Inertia = ?

m = mass of wheel = 2.3 kg

r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m

Therefore,

I = (2.3 kg)(0.25 m)²

I = 0.115 kg.m²

Now, we find the angular acceleration of the wheel:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s

ωi = Initial Angular Velocity = 0 rad/s

t = time = 5.5 s

Therefore,

α = (12.56 rad/s - 0 rad/s)/(5.5 s)

α = 2.28 rad/s²

Now, the torque is given as:

Torque = τ = Iα

τ = (0.115 kg.m²)(2.28 rad/s²)

τ = 0.26 N.m

Complete question:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?

A. the interior field points in a direction parallel to the exterior field

B. There is no electric field on the interior of the conducting sphere.

C. The interior field points in a direction perpendicular to the exterior field.

D. the interior field points in a direction opposite to the exterior field.

Answer:

B. There is no electric field on the interior of the conducting sphere.

Explanation:

Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).

Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of  free charges in the spherical conductor in response to any field until its neutralization.

Option B is correct.

There is no electric field on the interior of the conducting sphere.

Answer:

The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.

Explanation:

The energy stored in the inductor is given as

E₁ = ½LI²

The rate at which energy is stored in the inductor is

(dE₁/dt) = (d/dt) (½LI²)

Since L is a constant

(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)

(dE₁/dt) = LI (dI/dt)

Rate of Energy dissipated in a resistor = Power = I²R

(dE₂/dt) = I²R

When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field

(dE₁/dt) = (dE₂/dt)

OK (dI/dt) = I²R

L (dI/dt) = IR

Current in a this kind of series setup of inductor and resistor at any time, t, is given as

I = (V/R) (1 - e⁻ᵏᵗ)

k = (1/time constant) = (R/L)

(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ

L (dI/dt) = IR

L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)

V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)

e⁻ᵏᵗ = 1 - e⁻ᵏᵗ

2 e⁻ᵏᵗ = 1

e⁻ᵏᵗ = (1/2) = 0.5

e⁻ᵏᵗ = 0.5

In e⁻ᵏᵗ = In 0.5 = -0.69315

- kt = -0.69315

kt = 0.69315

k = (1/time constant)

Time constant = 36.0 ms = 0.036 s

k = (1/0.036) = 27.78

27.78t = 0.69315

t = (0.69315/27.78) = 0.02495 = 24.95 ms

Hope this Helps!!!

Answer:

A. 80.0microColoumbs

B.120.0 microcoloumbs

C.37.3v

Explanation:

See attached file

a. The charge Q1 on capacitor C1 should be 80.0 μC.

b. The charge on capacitor C3 should be 120 μC.

c. The applied voltage should be 37.5 v.

a. We know that

v = q/c

Q1/c1 = Q2/C2

Q1 = (C1/C2) Q2

= (6.00/3.00) * 40

= 80.0

b.

Q3 = Q1 + Q2

= (80 + 40)

= 120

c.

The voltage should be

= 80/ 6+ 120/5

= 37.5 V

hence,

a. The charge Q1 on capacitor C1 should be 80.0 μC.

b. The charge on capacitor C3 should be 120 μC.

c. The applied voltage should be 37.5 v.

Learn more about voltage here: https://brainly.com/question/14466758

Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = [tex]1 \times 10-12 W/m^2[/tex]

Now

Intensity level ( or Loudness)is

[tex]L = log10 \frac{I}{10}[/tex]

[tex]L = log10 \frac{10}{1\times 10^{-12}}[/tex]

[tex]L = log10 \times 1013[/tex]

[tex]= 13 \times 1 ( log10(10) = 1)[/tex]

Therefore  

L = 13 bel

And as we know that

1 bel = 10 decibels

So,

The  Sound intensity level is

L = 130 decibels

Answer:

In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

Explanation:

In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.

Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

Answer:

The wavelength is  [tex]\lambda = 6.28 *10^{-7}=628 nm[/tex]

The frequency is  [tex]f = 4.78 Hz[/tex]

Explanation:

From the question we are told that  

      The slit distance is [tex]d = 0.048 \ mm = 4.8 *0^{-5}\ m[/tex]

       The distance from the screen is  [tex]D = 6.50 \ m[/tex]

       The distance between fringes is  [tex]Y = 8.5 \ cm = 0.085 \ m[/tex]

Generally the distance between the fringes for a two slit interference is  mathematically represented as

           [tex]Y = \frac{\lambda * D}{d}[/tex]

=>       [tex]\lambda = \frac{Y * d }{D}[/tex]

substituting values      

           [tex]\lambda = \frac{0.085 * 4.8*10^{-5} }{6.50 }[/tex]

           [tex]\lambda = 6.27 *10^{-7}=628 nm[/tex]

Generally the frequency of the light is mathematically represented as

          [tex]f = \frac{c}{\lambda }[/tex]

where  c is  the speed of light with  values  

         [tex]c = 3.0 *10^{8} \ m/s[/tex]

substituting values  

      [tex]f = \frac{3.0*10^8}{6.28 *10^{-7}}[/tex]

      [tex]f = 4.78 Hz[/tex]