A compact disc (CD) is played by a cd player, which uses a laser to read the tracks on the disc. The disc spins initially at approximately 200 rotations per minute (rpm), and increases to a maximum of approximately 500 rpm as the laser spirals inward towards the center of the disc. This ensures that the laser covers equal distance in an equal amount of time during playback. During the time that the cd is played, which of the following statements is true?

a. The laser tracking mechanism experiences a changing tangential velocity
b. The laser tracking mechanism experiences a constant angular velocity
c. The laser tracking mechanism experiences a non-zero angular acceleration
d. The laser tracking mechanism experiences a non-zero tangential acceleration

Answer:

The Gas Giants have more moons.

Explanation:

Mercury-0

Venus-0

Earth-1

Mars-2

Jupiter-66

Saturn-62

Uranus-27

Neptune-13

Answer:

Half life is 3.23 hours

Explanation:

Given

Decay rate at starting = 1160 decays per minute

Decay rate after 4 hours = 170 decays per minute

As we know know

[tex]N = N_0 *e ^{\Lambda *T}[/tex]

Substituting the given values, we get -

[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]

Also

[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]

Substituting the given values we get -

[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

Hint : ΔV = Ed

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F[tex]_{net[/tex] = 0

so, F[tex]_E[/tex] = F[tex]_B[/tex]

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

Answer:

68 meters moved in the next seconds

Explanation:

Given

[tex]u= 30m/s[/tex]

[tex]a = 4m/s^2[/tex]

Required

Distance covered by the car in the next second

At a point in time t, the current distance is calculated as:

[tex]s(t) = ut + \frac{1}{2}at^2[/tex]

Substitute values for a and u in the above equation.

[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]

[tex]s(t) =30t + 2t^2[/tex]

Next, we generate the second degree Taylor polynomial as follows;

Calculate velocity (s'(t))

Differentiate s(t) to get velocity

[tex]s(t) =30t + 2t^2[/tex]

[tex]s'(t) =30 + 4t[/tex]

Calculate acceleration (s"(t))

Differentiate s'(t) to get acceleration

[tex]s'(t) =30 + 4t[/tex]

[tex]s"(t) =4[/tex]

When t = 0

We have:

[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]

[tex]s'(0) =30 + 4*0 = 30[/tex]

[tex]s"(0) = 4[/tex]

So, the second degree tailor series is:

[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]

To see the distance moved in the next second, we set t to 1

So, we have:

[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]

Solving s(1), s'(1) and s"(1)

We have:

[tex]s(1) =30*1 + 2*1^2 = 32[/tex]

[tex]s'(1) =30 + 4*1 = 34[/tex]

[tex]s"(1) =4[/tex]

Hence:

[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]

[tex]T_2(1) = 32 + 34 + 2[/tex]

[tex]T_2(1) = 68[/tex]

Answer:

When the spring compresses, elastic potential energy increases.

Answer:

the answer is b

Explanation:

elastic potential energy

Answer:

A boat travels for three hours with a... A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in still water?

Explanation:

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the concept of Logic.

Since in the above statement it is given that,

All S are P ==> True,

then obviously Some S are also P always, hence it is true.

Answer is True.

Answer:

conservation of linear momentum

We were told that two objects became stuck together hence we have to use the principle of conservation of  momentum to obtain the final velocities of the carts.

The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.

We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.

Learn more about conservation of momentum: https://brainly.com/question/11256472

Answer:

B)60km[W]

Explanation:

The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].

Hope that helps!

Answer:

 w = 0.319 rad / s

Explanation:

This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.

initial instant. Before the squirrel jumps

           L₀ = m v r

final instant. After the trough and the squirrel are together

          L_f = (I_fetter + I_ardilla) w

angular momentum is conserved

          L₀ = L_f

          m v r = (I_fetter + I_ardilla) w

          w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]

the moment inercial ofbody is

         I_thed = 2.00 kg m²

We approach the squirrel to a specific mass

          I_ardilla = m r²

we substitute

            w = m v r / ( I_[feefer  + m r²)

let's calculate

              w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )

              w = 0.6426 / 2.0119

               w = 0.319 rad / s

Answer:

C

Explanation:

Primacy means being first or important so thats not an important facial display as the others.

Answer:

d. the star is a member and also a part of an eclipsing binary star system.

Explanation:

If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).

The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.

Thus, option D is correct.

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

         Wₓ = 1200 sin 30 = 600 N

          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

        P + fr -Wx = 0

       fr = μ N

       fr = 0.20 1039.23

        fr = 207.85 N

we substitute

        P =  Wₓ -fr

        P = 600 - 207,846

        P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

         fr = 155.88 N

         P = 600 - 155.88

         P = 444.12 N

Answer:

θ = cos^(-1) (-A/B)

Explanation:

The image of the reauktant forces A & B are missing, so i have attached it.

Now, from the attached image, we will see that;

Angle between A and B is θ

Also;

A = Bcos(180° − θ)

Now, in trigonometry, we know that;

cos(180° − θ) = -cosθ

Thus;

A = -Bcosθ

cosθ = -A/B

Thus;

θ = cos^(-1) (-A/B)

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

[tex]A_1v_1 = A_2v_2[/tex]

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]

v₂ = 30 m/s

b)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]

m₁ = 12600 kg

c)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]

m₂ = 12600 kg

Answer:

mass, weight, strength of gravity increases, weight decreases

Explanation:

got it on edge

Answer:

An object’s

✔ mass

will remain constant throughout the universe, but its

✔ weight

can change from planet to planet.

If you increase the mass of a planet, what happens to its gravity?

✔ strength of gravity increases

If the gravity on a planet decreases, what happens to the weight of an object on that planet?

✔ weight decreases

Explanation:

right on edge 22

Answer:

soluble soluble soluble soluble

Explanation:

solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci

Answer:

C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.

Answer:

C.

A single nucleus divides into two or more nuclei and gives off energy.

hope it is helpful to you

Answer:

the smallest separation between two objects is 0.8067 m

Explanation:

Given the data in the question;

Altitude h = 5.75 km = 5750 m

Diameter D = 4.0 mm = 0.004 m

λ = 460 nm = 4.6 × 10⁻⁷ m

Now, Using Rayleigh criterion for Airy disks resolution.

we know that, Minimum angular separation for resolving two points is;

θ = 1.22λ / D

so we substitute

θ = (1.22 × 4.6 × 10⁻⁷)  / 0.004

θ = 5.612 × 10⁻⁷ / 0.004

θ = 1.403 × 10⁻⁴ rad  

so minimum separation [tex]d_{min[/tex] =  θh

so we substitute

[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m

[tex]d_{min[/tex] = 0.8067 m

Therefore, the smallest separation between two objects is 0.8067 m

Answer:

Solid:

- has definite shape and volume.

- change to liquid on heating.

Liquid:

- has definite volume but does not have definite shape .

- changes to solid on cooling.

Gas :

- does not have definite shape or volume.

- change to liquid on cooling

Answer:

50j

Explanation:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Answer:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Explanation:

Answer:

R = 7.915 10⁶ m,  M = 1.04 10³⁵ kg

Explanation:

Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations

           v = v₀ - g t

the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls

          v = -v₀

         -v₀ = v₀ - g t

          g = 2v₀ / t

          g = 2 15 / 2.7

          g = 11.11 m / s²

I now write the law of universal gravitation and Newton's second law

          F = m a

          G m M / R² = m a

         a = g

          g = G M / R²

Now let's work with the cruiser in orbit

         F = ma

acceleration is centripetal

         a = v² / r

         G m M / r² = m v² / r                        (1)

the distance from the center of the planet is

        r = R + h

        r = R + R = 2R

we substitute in 1

        G M / 4R² = v² / 2R

        G M / 2R = v²

The modulus of the velocity in a circular orbit is

         v = d / T

the distance is that of the circle

          d = 2π r

          v = 2π 2R / T

          v = 4π R / T

          G M / 2R = 16pi² R² / T²

          T² = 32 pi² R³ / GM

let's write the equations

             g = G M / R² (2)

             T² = 32 pi² R³ / GM

we have two equations and two unknowns, so it can be solved

let's clear the most on the planet and equalize

             g R² / G = 32 pi² R³ / GT²

              g T² = 32 pi² R

             R = g T² / 32 pi²

let's reduce the period to SI units

           T = 250 min (60 s / 1 min) = 1.5 104 s

let's calculate

             R = 11.11 (1.5 10⁴) ² / 32 π²

           R = 7.915 10⁶ m

from equation 2 we can find the mass of the planet

             M = g R² / G

             M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹

             M = 1.04 10³⁵ kg

Answer:

[tex]523269.9\ \text{N/m}[/tex]

Explanation:

q = Charge

r = Distance

[tex]q_1=25\ \text{C}[/tex]

[tex]r_1=3000\ \text{m}[/tex]

[tex]q_2=40\ \text{C}[/tex]

[tex]r_2=850\ \text{m}[/tex]

The electric field is given by

[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]

The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ([tex]U_{e}[/tex]), in joules, is equal to maximum translational kinetic energy ([tex]K[/tex]), in joules:

[tex]U_{e} = K[/tex]

[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)

Where:

[tex]k[/tex] - Spring constant, in newtons per meter.

[tex]A[/tex] - Amplitude, in meters.

[tex]m[/tex] - Object mass, in kilograms.

[tex]v[/tex] - Speed of the object at equilibrium, in meters per second.

If we know that [tex]k = 450\,\frac{N}{m}[/tex], [tex]m = 0.25\,kg[/tex] and [tex]v = 0.3\,\frac{m}{s}[/tex], then the amplitude of the motion is:

[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]k\cdot A^{2} = m\cdot v^{2}[/tex]

[tex]A = v\cdot \sqrt{\frac{m}{k} }[/tex]

[tex]A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }[/tex]

[tex]A \approx 0.274\,m[/tex]

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ([tex]E[/tex]), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ([tex]k = 450\,\frac{N}{m}[/tex], [tex]A \approx 0.274\,m[/tex])

[tex]E = U_{e}[/tex]

[tex]E = \frac{1}{2}\cdot k\cdot A^{2}[/tex]

[tex]E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}[/tex]

[tex]E = 16.892\,J[/tex]

The total energy of the object at any point of its motion is 16.892 joules.

Answer:

charge = electrons + protons

=92+92

=184

Newton’s first law of motion states that there must be a cause—which is a net external force—for there to be any change in velocity, either a change in magnitude or direction. An object sliding across a table or floor slows down due to the net force of friction acting on the object.

got it off g lol..

Answer:

it state that everybody in the universe is state that" universe continues its state of rest or uniform motion in a straight path unless it is acted upon by external force."

Answer:

I think it is C, I'm not for sure.

Answer:

0.02 kg/cm³

Explanation:

Answer:

Explanation:

energy of photon having wavelength of 400 nm = 1237.5/400 eV

= 3.1 eV.

Maximum kinetic energy of photoelectrons = 1.1 eV .

Threshold energy Ф = 3.1 - 1.1 = 2 eV .

energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .

Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy

= 4 - 2 = 2 eV .

Required kinetic energy K₀= 2 eV.

Answer:

surface

Explanation:

You need friction created by your tires and the road surface

to control your speed and direction.