If the diameter of a radar dish is doubled, what happens to its resolving power, assuming that all other factors remain unchanged? Its resolving power

a. is reduced to one-half its original value.
b. Quadruples.
c. is reduced to one-quartet its original value.
d. Halves.
e. Doubles.

Answers

Answer 1

Answer:

e. Doubles.

Explanation:

Resolving power is given by the formula as follows :

[tex]\dfrac{1}{d\theta}=\dfrac{D}{1.22\lambda}[/tex]

Here, [tex]d\theta[/tex] is the angle subtended by two distant objects

D is diameter of the telescope

Here, the diameter of a radar dish is doubled, assuming all other factors remain unchanged, then the resolving power gets doubled. Hence, the correct option is (e).


Related Questions

a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric field at the 30 cm mark

Answers

Answer:

-1748*10^N/C

Explanation:

See attached file

A tightly wound toroid of inner radius 1.2 cm and outer radius 2.4 cm has 960 turns of wire and carries a current of 2.5 A.

Requried:
a. What is the magnetic field at a distance of 0.9 cm from the center?
b. What is the field 1.2 cm from the center?

Answers

Answer:

a

  [tex]B = 0.0533 \ T[/tex]

b

  [tex]B = 0.04 \ T[/tex]

Explanation:

From the question we are told that

   The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]

   The  outer radius is  [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]

    The nu umber of turns is  [tex]N = 960[/tex]

    The current it is carrying is  [tex]I = 2. 5 A[/tex]

Generally the magnetic field is mathematically represented as

      [tex]B = \frac{\mu_o * N* I }{2 * \pi * r }[/tex]

Where  [tex]\mu_o[/tex] is the permeability of free space with a constant value    

            [tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]

And the given distance where the magnetic field is felt is  r =  0.9 cm  =  0.009 m

Now  substituting values

     [tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]

    [tex]B = 0.0533 \ T[/tex]

    Fro the second question the distance of the position considered from the center is  r =  1.2 cm  =  0.012 m

So the  magnetic field is  

        [tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]

        [tex]B = 0.04 \ T[/tex]

The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.

The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.

The given parameters;

radius of the toroid, r = 1.2 cm = 0.012 mouter radius of the toroid, R = 2.4 cm = 0.024 mnumber of turns, N = 960 turnscurrent in wire, I = 2.5 A

The magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;

[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]

The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;

[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]

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find the value of k for which the given pair of vectors are not equal
2ki +3j​ and 8i + 4kj

Answers

Answer:

5

Explanation:

A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is µs = 0.5, how far from the axis of rotation can he stand without sliding?

Answers

Answer:

0.8 m

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing towards the center.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the centripetal direction:

∑F = ma

Nμ = m v²/r

Substitute and simplify:

mgμ = m v²/r

gμ = v²/r

Write v in terms of ω and solve for r:

gμ = ω²r

r = gμ/ω²

Plug in values:

r = (10 m/s²) (0.5) / (2.5 rad/s)²

r = 0.8 m

The distance (radius) from the axis of rotation which the man can stand without sliding is 0.784 meters.

Given the following data:

Angular speed = 2.5 rad/s.Coefficient of static friction = 0.5

To determine how far (radius) from the axis of rotation can the man stand without sliding:

We would apply Newton's Second Law of Motion, to express the centripetal and force of static friction acting on the man.

[tex]\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n[/tex]....equation 1.

But, Normal force, [tex]F_n = mg[/tex]  

Substituting the normal force into eqn. 1, we have:

[tex]\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug[/tex]....equation 2.

Also, Linear speed, [tex]v = r\omega[/tex]

Substituting Linear speed into eqn. 2, we have:

[tex]\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}[/tex]

Radius, r = 0.784 meters

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Consider a sound wave modeled with the equation s(x, t) = 3.00 nm cos(3.50 m−1x − 1,800 s−1t). What is the maximum displacement (in nm), the wavelength (in m), the frequency (in Hz), and the speed (in m/s) of the sound wave?

Answers

Answer:

-   maximum displacement = 3.00nm

-   λ = 1.79m

-  f = 286.47 s^-1

Explanation:

You have the following equation for a sound wave:

[tex]s(x,t)=3.00nm\ cos(3.50m^{-1}x- 1,800s^{-1} t)[/tex]              (1)

The general form of the equation of a sound wave can be expressed as the following formula:

[tex]s(x,t)=Acos(kx-\omega t)[/tex]            (2)

A: amplitude of the wave = 3.00nm

k: wave number = 3.50m^-1

w: angular frequency = 1,800s^-1

- The maximum displacement of the wave is given by the amplitude of the wave, then you have:

maximum displacement = A = 3.00nm

- The wavelength is given by :

[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{3.50m^{-1}}=1.79m[/tex]

The values for the wavelength is 1.79m

- The frequency is:

[tex]f=\frac{\omega}{2\pi}=\frac{1,800s^{-1}}{2\pi}=286.47s^{-1}[/tex]

The frequency is 286.47s-1

Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer

Answers

Answer:

Total heat transfer is positive

Total work transfer is positive

Explanation:

The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.

Heat transfer to a system is positive and that transferred from the system is negative.

Also, work done by a system is positive while the work done on the system is negative.

Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)

Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos(t) + 9 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.

Answers

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Calculate the ideal banking angle in degrees for a gentle turn of 1.88 km radius on a highway with a 136.3 km/hr speed limit, assuming everyone travels at the speed limit.

Answers

Answer:

Ф = 4.4°

Explanation:

given:

radius (r) = 1.88 km

velocity (v) = 136.3 km/hr

required:

banking angle ∡ ?

first:

convert 1.88 km to m = 1.88km * 1000m / 1km

r = 1880 m

convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)

v = 37.86 m/s

now.. calculate the angle

Ф = inv tan (v² / r * g)            we know that gravity = 9.8 m/s²

Ф = inv tan (37.86² / (1880 * 9.8))

Ф = 4.4°

Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.

Answers

Answer:

Objects act differently in a gravity field than in an accelerating reference frame.

Explanation:

The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.

The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.

Answer:

A - The orbital path of mercury around the sun has changed.

Explanation:

got right on edg.

What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)

Answers

Answer:

ΔU = 2.25 x 10⁸ J

t = 104.17 s

Explanation:

The change in internal energy of the engine can be given by the following formula:

ΔU = (Mass of Gasoline)(Energy Content of Gasoline)

ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)

ΔU = 2.25 x 10⁸ J

Now, for the time of operation, we use the following formula of power.

P = W/t = ΔU/t

t = ΔU/P

where,

t = time of operation = ?

ΔU = Change in internal energy = 2.25 x 10⁸ J

P = Power of motor = 600 W

Therefore,

t = (2.25 x 10⁸ J)/(600 W)

t = (375000 s)(1 h/3600 s)

t = 104.17 s

2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the

Answers

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

   

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage

Answers

Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

Can an object travel at the speed of
light? Why or why nbt?

Answers

Answer:

no the only things that can travel at the speed of light are waves in the electromagnetic spectrum

No because, the object shrinks as it moves forward compared to light, it never does.

Source: google

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c

Answers

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c

What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.

Answers

Answer:

[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]

Explanation:

Assume the stone consists of basalt, which has a density of 3.0 g/cm³.

[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]

6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.​

Answers

Answer:

same direction = 80 (n)

opposite direction = 20 (n) going one direction

Explanation:

same direction means they are added to each other

and opposite means acting on eachother

A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t​, where v (t )is in meters per second. ​a) How far does the particle travel during the first 2 ​sec? ​b) How far does it travel during the second 2 ​sec?

Answers

Answer:

a) 2.933 m

b) 4.534 m

Explanation:

We're given the equation

v(t) = -0.4t² + 2t

If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.

See attachment for the calculations

The conclusion of the attachment will be

7.467 - 2.933 and that is 4.534 m

Thus, The distance it travels in the second 2 sec is 4.534 m

A medieval city has the shape of a square and is protected by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round your answers to one decimal place. Use g ≈ 9.8 m/s2. Enter your answer using interval notation. Enter your answer in terms of degrees without using a degree symbol.)

Answers

Answer:

  θ₁ = 85.5º       θ₂ = 12.98º

Explanation:

Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.

Let's write the equations for motion for this point

X axis

          x = v₀ₓ t

          x = v₀ cos θ t

Y axis

         y = [tex]v_{oy}[/tex] t - ½ g t2

         y = v_{o} sin θ t - ½ g t²

let's substitute the values

         100 = 80 cos θ t

           15 = 80 sin θ t - ½ 9.8 t²

we have two equations with two unknowns, so the system can be solved

let's clear the time in the first equation

           t = 100/80 cos θ

         15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²

         15 = 100  tan θ - 7.656 sec² θ

we can use the trigonometric relationship

         sec² θ = 1- tan² θ

we substitute

       15 = 100 tan θ - 7,656 (1- tan² θ)

       15 = 100 tan θ - 7,656 + 7,656 tan² θ

        7,656 tan² θ + 100 tan θ -22,656=0

let's change variables

       tan θ = u

         

        u² + 13.06 u + 2,959 = 0

let's solve the quadratic equation

       u = [-13.06 ±√(13.06² - 4  2,959)] / 2

       u = [13.06 ± 12.599] / 2

        u₁ = 12.8295

        u₂ = 0.2305

now we can find the angles

         u = tan θ

         θ = tan⁻¹ u

        θ₁ = 85.5º

         θ₂ = 12.98º

A guitar string 0.65 m long has a tension of 61 N and a mass per unit length of 3.0 g/m. (i) What is the speed of waves on the string when it is plucked? (ii) What is the string's fundamental frequency of vibration when plucked? (iii) At what other frequencies will this string vibrate?

Answers

Answer:

i

  [tex]v = 142.595 \ m/s[/tex]

ii

  [tex]f = 109.69 \ Hz[/tex]

iii1 )

  [tex]f_2 =219.4 Hz[/tex]

iii2)

   [tex]f_3 =329.1 Hz[/tex]

iii3)

    [tex]f_4 =438.8 Hz[/tex]

Explanation:

From the question we are told that

    The length of the string is  [tex]l = 0.65 \ m[/tex]

     The tension on the string is  [tex]T = 61 \ N[/tex]

     The mass per unit length is  [tex]m = 3.0 \ g/m = 3.0 * \frac{1}{1000} = 3 *10^{-3 } \ kg /m[/tex]

     

The speed of wave on the string is mathematically represented as

       [tex]v = \sqrt{\frac{T}{m} }[/tex]

substituting values

      [tex]v = \sqrt{\frac{61}{3*10^{-3}} }[/tex]

     [tex]v = 142.595 \ m/s[/tex]

generally the  string's  frequency is mathematically represented as

         [tex]f = \frac{nv}{2l}[/tex]

n = 1  given that the frequency we are to find is the fundamental frequency

So

      substituting values

       [tex]f = \frac{142.595 * 1 }{2 * 0.65}[/tex]

       [tex]f = 109.69 \ Hz[/tex]

The  frequencies at which the string would vibrate include

1       [tex]f_2 = 2 * f[/tex]

Here [tex]f_2[/tex] is  know as the second harmonic and the value is  

      [tex]f_2 = 2 * 109.69[/tex]

      [tex]f_2 =219.4 Hz[/tex]

2

[tex]f_3 = 3 * f[/tex]

Here [tex]f_3[/tex] is  know as the third harmonic and the value is  

      [tex]f_3 = 3 * 109.69[/tex]

     [tex]f_3 =329.1 Hz[/tex]

3

     [tex]f_3 = 4 * f[/tex]

Here [tex]f_4[/tex] is  know as the fourth harmonic and the value is  

      [tex]f_3 = 4 * 109.69[/tex]

     [tex]f_4 =438.8 Hz[/tex]

The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period

Answers

Answer:

Q = 23.49 C

Explanation:

We have,

Electric current drawn by the air conditioner is 16.2 A

Time, t = 1.45 s

It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,

[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]

So, the charge of 23.49 C is passing through the circuit during this period.

The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Answers

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)

The electric field on the sphere is also given as;

[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]

[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]

Substitute in the value of q in equation (1)

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]

Therefore, the potential at the center of the sphere is -924 V

Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .

Answers

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.

What are Velocity and Acceleration?

The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.

There are several types of velocity :

Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform Velocity

The pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.

According to the question, the given values are :

Time, t = 3.50 sec

Initial Velocity, u = 0 m/s

Use equation of motion :

v = u+at

v = 0+ g sin 18 × 3.5

v = 10.6 m/s.

So, the final velocity will be 10.6 m/s.

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Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muons are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?

Answers

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.  

An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about its vertical axis. Riders stand inside against the carpeted surface and rotate with the cylinder while it accelerates to its full angular velocity. At that point the floor drops away and friction between the riders and the cylinder prevents them from sliding downward. The coefficient of static friction between the riders and the cylinder is 0.87. What minimum angular velocity in radians/second is necessary to assure that the riders will not slide down the wall?

Answers

Answer:

The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.

Explanation:

The riders will experience a centripetal force from the cylinder

[tex]F_{C}[/tex] = mrω^2    .... equ 1

where

m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of of the rider

For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as

[tex]F_{f}[/tex] = μR       ....equ 2

where

μ = coefficient of friction = 0.87

R is the normal force from the rider = mg

where

m is the rider's mass

g is the acceleration due to gravity = 9.81 m/s

substitute mg for R in equ 2, we'll have

[tex]F_{f}[/tex] = μmg     ....equ 3

Equating centripetal force of equ 1 and frictional force of equ 3, we'll get

mrω^2 = μmg

the mass of the rider cancels out, and we are left with

rω^2 = μg

ω^2 = μg/r

ω = [tex]\sqrt{\frac{ug}{r} }[/tex]

ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]

ω = 1.58 rad/second

The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s

The riders will experience a  centripetal force from the cylinder

[tex]F = mrw^2[/tex]

where  m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of the rider

For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as

f = μN

where

μ = coefficient of friction = 0.87

N is the normal force = mg

f = μmg  

Equating centripetal force of and frictional force of we'll get

[tex]mrw^2 = umg[/tex]

[tex]rw^2 = ug[/tex]

[tex]w^2 = ug/r[/tex]

[tex]w= \sqrt{ug/r}[/tex]

[tex]w= \sqrt{0.87*9.8/3.4}[/tex]  

ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.

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A length of organ pipe is closed at one end. If the speed of sound is 344 m/s, what length of pipe (in cm) is needed to obtain a fundamental frequency of 50 Hz

Answers

Answer:

The length = 27.52m

Explanation:

v=f x wavelength

what is drift speed ? {electricity}​

Answers

Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

Explanation:

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

Answers

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.

_____________ friction is the interlocking of surfaces due to irregularities on the surfaces preventing those surfaces from moving/sliding against each other. For surfaces moving/sliding on each other, ___________ friction overwhelms kinetic friction to that movement/sliding. Kinetic friction is alway larger than ____________ friction. Kinetic friction is alway equal to _________ friction.

Answers

Answer:

STATIC,  STATIC

KINETIC friction is less than static friction

Explanation:

In this exercise you are asked to complete the sentences with the correct words.

STATIC friction prevents the relative movement of two surfaces in contact.

For moving surfaces the friction is STATIC is greater than the kinetic friction.

For the last two sentences I think they are misspelled, the correct thing is

KINETIC friction is less than static friction

what is the preferred method of using percentage data by using a circle divided into sections

Answers

Answer:

A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole

Explanation:

pie charts are a useful way to organize data in order to see the size of components relative to the whole.

Consider a conducting rod of length 31 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails. At what speed (in m /s) must the sliding rod move to produce an emf of 0.75 V in a 1.75 T field?

Answers

Answer:

The speed of the rod is 1.383 m/s

Explanation:

Given;

length of the conducting rod, L = 31 cm = 0.31 m

induced emf on the rod, emf = 0.75V

magnetic field around the rod, B = 1.75 T

Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.

emef = BLv

where;

B is the magnetic field

L is length of the rod

v is the speed of the rod

v = emf / BL

v = (0.75) / (1.75 x 0.31)

v = 1.383 m/s

Therefore, the speed of the rod is 1.383 m/s

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