Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.

Given.

force = 16.88 N is

radius = 0.340m

an angular acceleration = 1.20rad/s^2

the formula for torque is

F*r = I*a

where I is moment of inertia

16.88*.34 = I*1.2

I = 4.78Kg-m^2

so rotational inertia I = 4.78Kg-m^2

Answer:

As in explanation.

Explanation:

A) Centre of Curvature: This is defined as the point in the center of the sphere from which the mirror was sliced. It is represented by the letter "C"

B) Vertex: It is defined as the point on the mirror's surface where the principal axis meets the mirror. It is represented by the letter A.

C) Focal Point: This is defined as the Midway point between the vertex and the center of curvature. It is represented by the letter "F"

D) Radius of Curvature: This is defined as the distance from the vertex to the center of curvature. It is represented by the letter "R"

E) Focal Length: This is defined as the distance from the mirror to the focal point. It's represented by the letter "f"

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Answer:

Answer in explanation

Explanation:

Water is mainly used as coolant in heating systems like hot-water radiators. The main function of water in such systems, is to absorb as much heat as possible, in order to decrease the temperature of the system and as a result cool it.

The specific heat capacity is the measure of heat energy that is required to raise the temperature of unit mass of a substance through 1 °C. In other words, specific heat capacity quantifies the amount of heat that can be stored by a unit mass of a substance having a degree rise in temperature.

Thus, the more specific heat a substance has, the more heat it can absorb from the hot system. Hence, the specific heat capacity of a coolant must be high.

This is the reason why water, with its high specific heat capacity, is utilized for heating systems, such as radiators.

Answer:

variable

Explanation:

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω ,  R₃ = 1.72 10³ Ω ,  R₄ = 5.4 10³ Ω    1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

   V = i R

let the voltage

1st resistance

         V = i R

         R₁ = V / i

         R₁ = 14.7 / 1 10⁻³

         R₁ = 14.7 10³ Ω

power is

        P = V i

        P = 14.7 1 10⁻³

        P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

          R₂ = 8.18 / 1 10⁻³

          R₂ = 8.18 10³ Ω

Power

          P = 8.18 1 10⁻³

          P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

          V₁ + V₂ + V₃ = 24.6

          V₃ = 24.6 - V₁ -V₂

          V₃ = 24.6 - 14.7 - 8.18

          V₃ = 1.72 V

          R₃ = 1.72 / 1 10⁻³

          R₃ = 1.72 10³ Ω

power

          P = Vi

          P = 1.72 10⁻³

          P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

           30 = V₁ + V₂ + V₃ + V₄

           V₄ = 30 - V₁ -V₂ -V₃

           V₄ = 30 -14.7 - 8.18 - 1.72

           V₄ = 5.4 V

          R₄ = 5.4 / 1 10⁻³

          R₄ = 5.4 10³ Ω

Power

         P = V i

         P = 5.4 10⁻³

         P = 0.0054 W

⅛ W resistance

The values ​​of these resistance are commercially

Let's check the consumption of the circuit

  R_total = R₁ + R₂ + R₃ + R₄

  R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

   R_total = 30 10³

the current circulating in the circuit is

     i = V / R_total

     i = 30/30 10³

     i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

Answer:

constructive interferencia  0, 1 , 2 m

destructive inteferencia   1/4, 3/4. 5/4, 7/4 m

Explanation:

This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression

            Δr / λ = Φ / 2π

            Δr = Φ/2π   λ

let's apply this expression to our case

λ = 1 m

            Δr = Φ 1 / 2π

We have constructive interference for angle of  Φ = 0, 2π, ...

let's find the values ​​where they occur

  Φ         Δr

   0          0

  2π         1

  4π        2

Destructive interference occurs by    Φ = π /2, 3π / 2, ...

 Φ          Δr

 π/2       ¼ m

 3π /2    ¾ m

5π /2     5/4 m

7π /2      7/4 m

Answer:

y = 1/2at^2

we could also write it as-

y = (at^2)/2

2y = at^2

2y/a = t^2

√2y/a = t

hope it helps

Answer:

Explanation:

Answer:

quantum entanglement is thought to be one of the trickiest concepts in science, but the core issues are simple. And once understood, entanglement opens up a richer understanding of concepts such as the “many worlds” of quantum theory.

Explanation:

Answer:

The tension in the cord provides by centripetal force

T = Fc

= mv^2/r

= 2kg ( 6)^2/2

=36 N

Answer:

1.6×10²⁰

Explanation:

An ampere is a Coulomb per second.

1 A = 1 C / s

The amount of charge after 5 seconds is:

5.0 A × 5 s = 25 C

The number of electrons is:

25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons

Answer:

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Explanation:

given data

frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz

we consider here intensity I = 0.2 W/m²

solution

we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s

and take energy density is E

so here

E × C = I  

E = [tex]\frac{I}{C}[/tex]   ................1

here C = 3 × [tex]10^{8}[/tex] m/s

so photon density is

photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex]     ...............2

photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Answer:

   t = t₀ / 2

Explanation:

In this exercise we must use Newton's second law

          F = m a

          a = F / m

now we can use kinematics

  as in object part of rest (v₀ = 0)

        v =a t₀

        t₀ = v / a

these results are with the first experiment

now repeat the experiment, but F = 2F₀

           a = 2F₀ / m = 2 a₀

          v = 2 a₀ t

          t = v / 2a₀

          t = t₀ / 2

The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].

Given the following data:

To find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:

Mathematically, Newton's Second Law of Motion is given by this formula;

[tex]F = \frac{M(V-U)}{t}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]

When the experiment is repeated, the magnitude of the force is doubled:

[tex]F = 2F[/tex]

Now, we can find the time interval that is required to reach the same final speed (V):

[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]

Substituting the value of F, we have:

[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]

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Answer:

The separation of the two slits is 0.456 mm.

Explanation:

Given the wavelength of light = 519 nm

The indifference pattern = 4.6 m

Adjacent bright fringes = 5.2 mm

In the interference, the equation required is Y = mLR/d

Here, d sin theta = mL

L = wavelgnth

For bright bands, m is the  order = 1,2,3,4  

For dark bands,  m = 1.5, 2.5, 3.5, 4.5

R = Distance from slit to screen (The indifference pattern)

Y = Distance from central spot to the nth  order fringe or fringe width

Thus,  here d = mLR/Y

d = 1× 519nm × 4.6 / 5.2mm

d = 0.459 mm

Answer:

The new time period is  [tex]T_2 = 3.8 \ s[/tex]

Explanation:

From the question we are told that

  The period of oscillation is  [tex]T = 5 \ s[/tex]

   The  new  length is  [tex]l_2 = 0.76 \ m[/tex]

Let assume the original length was [tex]l_1 = 1 m[/tex]

Generally the time period is mathematically represented as

         [tex]T = 2 \pi \sqrt{ \frac{ I }{ mgh } }[/tex]

Now  I is the moment of inertia of the stick which is mathematically represented as

           [tex]I = \frac{m * l^2 }{12 }[/tex]

So

        [tex]T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }[/tex]

Looking at the above equation we see that

        [tex]T \ \ \ \alpha \ \ \ l[/tex]

=>    [tex]\frac{ T_2 }{T_1} = \frac{l_2}{l_1}[/tex]

=>    [tex]\frac{ T_2}{5} = \frac{0.76}{1}[/tex]

=>     [tex]T_2 = 3.8 \ s[/tex]

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

Answer:

A. 0.0374C

B. 0.012F

C. 18 ohms

Explanation:

See attached file

Answer:

Explanation:

Using the lens formula

1//f = 1/u+1/v

f is the focal length of the lens

u is the object distance

v is the image distance

For convex lens

The focal length of a convex lens is positive and the image distance can either be negative or positive.

Given f = 20cm and u = 10cm

1/v = 1/f - 1/u

1/v = 1/20-1/10

1/v = (1-2)/20

1/V = -1/20

v = -20/1

v = -20 cm

Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image

For concave lens

The focal length of a concave lens is negative and the image distance is negative.

Given f = -20cm and u = 10cm

1/v = 1/f - 1/u

1/v = -1/20-1/10

1/v = (-1-2)/20

1/V = -3/20

v = -20/3

v = -6.67 cm

Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".

The region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.

According to the question,

Current, I = 2.4 A

Radius, R = 5 cm or,

                = 0.05 m

Constant, μ₀ = 4π × 10⁻⁷ H/m

We know the relation,

The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]

By substituting the values in the above relation, we get

                                    = [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]

                                    = 9.6 × 10⁻⁴ T

Thus the above answer is appropriate.

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Answer:

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

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Answer:

The magnification is  [tex]m = 12[/tex]

Explanation:

From the question  we are told that

   The object distance is [tex]u = 36.2 \ cm[/tex]

     The focal length is  [tex]v = 39.5 \ cm[/tex]

From the lens equation we have that

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

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

substituting values

       [tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]

       [tex]\frac{1}{v} = -0.0023[/tex]

=>   [tex]v = \frac{1}{0.0023}[/tex]

=>   [tex]v =-433.3 \ cm[/tex]

The magnification is mathematically represented as

         [tex]m =- \frac{v}{u}[/tex]

substituting values

        [tex]m =- \frac{-433.3}{36.2}[/tex]

         [tex]m = 12[/tex]

Answer:

The amplitude [tex]A(5) = 1 \ m[/tex]

Explanation:

From the question we are told that

     The  spring constant is  [tex]k = 100 \ N/m[/tex]

      The  damping constant is  [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]

       The mass is  [tex]m = 0.050 \ kg[/tex]

       The  maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]

       The  time  considered is  t =  5.0 s

Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as

           [tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]

substituting values

         [tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]

         [tex]A(5) = 1 \ m[/tex]

Answer:

Hey there!

PE=mgh, so as height decreases, so does the potential energy.

KE=mv^2, so as velocity increases, kinetic energy increases.

Thus, the correct answer would be Decreases, Increases.

Let me know if this helps :)

Given that,

Central maximum = 1 cm

Distance from the window shade to the wall =4 m

We know that,

The visible range of the sun light is 400 nm to 700 nm.

(a). We need to calculate the average wavelength

Using formula of average wavelength

[tex]\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}[/tex]

Put the value into the formula

[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]

[tex]\lambda_{avg}=550\ nm[/tex]

(b). We need to calculate the diameter of the pinhole

Using formula for diameter

[tex]w=\dfrac{2.44\lambda L}{D}[/tex]

[tex]D=\dfrac{2.44\lambda L}{w}[/tex]

Put the value into the formula

[tex]D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}[/tex]

[tex]D=0.537\ mm[/tex]

Hence, (a). The average wavelength 550 nm.

(b). The diameter of the pinhole is 0.537 mm.

Answer:

10.75 A

The current is in opposite direction since it causes a repulsion force between the wires

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]

where [tex]F/l[/tex] is the force per unit length on the wires

[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A

[tex]I_{1}[/tex] = current on the first wire = 0.520 A

[tex]I_{2}[/tex] = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] =  [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]

4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]

[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A

The current is in opposite direction since it causes a repulsion force between the wires.

Answer:

It will be half that if the first wave

Explanation:

Because the wave speed remains the same, the result of doubling the frequency is that the wavelength is half as large as it

Answer:

there are over 100 billion stars in our galaxy.