A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative velocity of 0.40c. What is the frequency of the beacon radio signal that is detected on the ship? Answer in MHz with no decimal places.
A) 55 MHz
B) 60 MHz
C) 66 MHz
D) 71 MHz
E) 76 MHz

Answers

Answer 1

Answer:

To first order  f' = f (1 - u/c) where observer and source are separating

f' = (1 + .4) = 70 MHz   so (d) would be correct

The next term would be 1/2 (u/c)^2 = 1/2 * ,4^2 = .08 and the correction would be .08 * 50 = 4 Mhz

Most approximations would use (d).


Related Questions

please i dont understand

Answers

Give me points I will help you

In a physics lab, Asha is given a 11.5 kg uniform rectangular plate with edge lengths 62.9 cm by 46.9 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .

Answers

Answer:

6.9kgm²

Explanation:

For an axis through the center of the rectangle, I = m[(w²+L²)/12

Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]

Adding the 2 expressions,

I = (m/3)*(w²+L²)

I =6.95 kg∙m²

Which of the units of the following physical quantities are derived

Answers

Answer:

where is the attachment

Explanation:

1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

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

Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?

Answers

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

Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided

Answers

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

Using this information...
Determine the velocity of the pebble as it passes over the top of the tree.​

Answers

[tex]19.2\:\text{m/s}[/tex]

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

[tex]v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)[/tex]

[tex]\:\:\:\:\:= 19.2\:\text{m/s}[/tex]

The height of the tree is approximately 12.5 meters when velocity of the pebble as it passes over the top of the tree.​

Let's calculate the height of the tree step by step:

Given:

Initial velocity (v0) = 25 m/s

Launch angle (θ) = 40° above the horizontal

Time after launch (t) = 2 seconds

Acceleration due to gravity (g) = -9.8 m/s² (negative because it acts downward)

Step 1: Calculate the vertical component of the initial velocity (Vy):

Vy = v0 * sin(θ)

Vy = 25 m/s * sin(40°)

Vy ≈ 25 m/s * 0.6428 ≈ 16.07 m/s (rounded off to two decimal places)

Step 2: Calculate the vertical displacement (change in height) of the pebble after 2 seconds:

d = vot + (1/2)at²

d = (16.07 m/s) * (2 s) + (1/2) * (-9.8 m/s²) * (2 s)²

d ≈ 32.14 m - 19.6 m

d ≈ 12.54 m (rounded off to two decimal places)

Step 3: The height of the tree is equal to the vertical displacement of the pebble:

Height of the tree ≈ 12.54 m ≈ 12.5 m (rounded off to one decimal place)

The height of the tree is approximately 12.5 meters.

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Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

Answers

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".

Magnetic field

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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Define the following, and give the letter which we will abbreviate them by:
Center of curvature:
Vertex:
Focal Point:
Radius of curvature:
Focal length:

Answers

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"

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?

Answers

Answer:

At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.

Explanation:

The number of particles (N)  counts are inversely proportional to the distance between the source and the detector.  

By using the below formula we can find the number of counts.

[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]

The number of count per second, when the distance is 3 meters.

[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]

Number of count per second when the distance is 10 meters.

[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]

A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.

a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above

Answers

Answer:

None of the above

Explanation:

The formula of the magnetic field of a point next to a wire with current is:

B = 2×10^(-7) × ( I /d)

I is the intensity of the current.

d is the distance between the wire and the point.

● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T

Explain why water, with its high specific heat capacity, is utilized for heating systems such as hot-water radiators.

Answers

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.

A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (s ^ 2) . Calculate the rotational inertia of the wheel. A. 2.77 kg - m ^ 2 B. 0.73 kg - m ^ 2 C. 4.41 kg - m ^ 2 O. 4.78 kg - m ^ 2

Answers

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

All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outlet. (The three devices are in parallel when plugged into the same outlet.) Assume that this is the standard household 120-V circuit with a 15-A fuse.
a. What current is drawn by each device?
b. To see if this combination will blow the 15-A fuse, find the total current used when all three appliances are on.

Answers

Answer:

total current = 12.417 A

so it will not fuse as current is less than 15 A

Explanation:

given data

toaster = 1140-W

blender = 270-W

lamp = 80-W

voltage = 120 V

solution

we know that current is express as

current = power ÷ voltage   ......................1

here voltage is same in all three device

so

current by toaster is

I = [tex]\frac{1140}{120}[/tex]

I = 9.5 A

and

current by blender

I = [tex]\frac{270}{120}[/tex]

I = 2.25 A

and

current by lamp is

I = [tex]\frac{80}{120}[/tex]

I = 0.667 A

so here device in parallel so

total current is = 9.5 A + 2.25 A + 0.667 A

total current = 12.417 A

so it will not fuse as current is less than 15 A

An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?

Answers

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

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification

Answers

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]

         

a radio antenna emits electromagnetic waves at a frequency of 100 mhz and intensity of what is the photon density

Answers

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

What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?​

Answers

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

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a 55m wide tub is 0.80 Hz.

Required:
What is the speed of the water wave?

Answers

Answer:

The  speed of the water wave is [tex]v = 88 \ m/s[/tex]

Explanation:

From the question we are told that

      The  width of the tube is  [tex]L = 55 \ m[/tex]

     The fundamental  frequency is  [tex]f = 0.80 \ Hz[/tex]

Generally the fundamental frequency is mathematically represented as

      [tex]f = \frac{v}{2 * L }[/tex]

=>    [tex]v = f * 2 * L[/tex]

substituting values

       [tex]v = 0.8 * 2 * 55[/tex]

       [tex]v = 88 \ m/s[/tex]

The speed of the water wave will be 88 m/s.

Given information:

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center.

The frequency of the standing wave is [tex]f=0.8[/tex] Hz.

The width of the tub is [tex]w=55[/tex] m.

Let v be the speed of the standing wave.

The speed of the wave can be calculated as,

[tex]v=2wf\\v=2\times 55\times 0.8\\v=88\rm\; m/s[/tex]

Therefore, the speed of the water wave will be 88 m/s.

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A rock weighing 20 N (mass = 2 kg) is swung in a horizontal circle of radius 2 m at a constant speed of 6 m/s. What is the tension in the cord?

Answers

Answer:

The tension in the cord provides by centripetal force

T = Fc

= mv^2/r

= 2kg ( 6)^2/2

=36 N

Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?

a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100

Answers

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

Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.

Answers

Answer:

137.69°

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°

Phase angle:

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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A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?

Answers

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]

       

As a skydiver falls, his potential energy ___ and his kinetic energy __​
increases,increases
increases,decreases
decreases,increases
decreases, decreases

Answers

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 :)

M
A boy of mass 60 kg and a girl of mass 40 kg are
together and at rest on a frozen pond and push
each other apart. The girl moves in a negative
direction with a speed of 3 m/s. What must be the
total final momentum of the boy AND girl
combined?
A. -120 kgm/s
B. 0 kgm/s
C. -100 kgm/s
D. 120 kgm/s

Answers

Answer:

option D thinking so

Explanation:

okay na your whish

If you were to come back to our solar system in 6 billion years, what might you expect to find?

A) a red giant star

B) a rapidly spinning pulsar

C) a white dwarf

D) a black hole

E) Everything will be essentially the same as it is now

Answers

Answer:

A)a red giant star

a. The molecules of a magnet are independent...​

Answers

Answer:

variable

Explanation:

At what speed, as a fraction of c, will a moving rod have a length 65% that of an identical rod at rest

Answers

Answer:

v/c = 0.76

Explanation:

Formula for Length contraction is given by;

L = L_o(√(1 - (v²/c²))

Where;

L is the length of the object at a moving speed v

L_o is the length of the object at rest

v is the speed of the object

c is speed of light

Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.

Thus;

0.65L_o = L_o[√(1 - (v²/c²))]

Dividing both sides by L_o gives;

0.65 = √(1 - (v²/c²))

Squaring both sides, we have;

0.65² = (1 - (v²/c²))

v²/c² = 1 - 0.65²

v²/c² = 0.5775

Taking square root of both sides gives;

v/c = 0.76

Four friends push on the same block in different directions. Allie pushes on the block to the north with a force of 18 N. Bill pushes on the block to the east with a force of 14 N. Chris pushes on the block to south with a force of 23 N. Debra pushes on the block to the west with a force of 20 N. Assuming it does not move vertically, in which directions will the block move? north and west south and east south and west north and east

Answers

Answer:

South and West

Explanation:

Those people are pushing the hardest. It will move south faster than it moves west.

Are Quantum Physics, Quantum mechanics,Quantum Engagement same?
or, Do they branch of each others ​

Answers

Answer:

The topic of quantum entanglement is at the heart of the disparity between classical and quantum physics: entanglement is a primary feature of quantum mechanics lacking in classical mechanics. ... In the case of entangled particles, such a measurement will affect the entangled system as a whole

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:

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