Terms to describe the opposition by a material.to being magnetised is

Answers

Answer 1

Answer:

Repulsion

Explanation:


Related Questions

The highest mountain on mars is olympus mons, rising 22000 meters above the martian surface. If we were to throw an object horizontaly off the mountain top, how long would it take to reach the surface? (Ignore atmospheric drag forces and use gMars=3.72m/s^2

a. 2.4 minutes
b. 0.79 minutes
c. 1.8 minutes
d. 3.0 minutes

Answers

Answer:

  t = 1.81 min ,     the correct answer is c

Explanation:

This is a missile throwing exercise

The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation

             y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

the final height is y = 0 and the initial height is y₀ = 22000 m

            0 = y₀ + 0 - ½ g t²

             

            t = √y 2y₀ / g

let's calculate

           t = √(2  22000 / 3.72)

           t = 108.76 s

let's reduce to minutes

           t = 108.76 s (1 min / 60 s)

           t = 1.81 min

The correct answer is c

In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?

Answers

Answer:

 Δy = 1 10⁻⁴ m

Explanation:

In double-slit experiments the constructive interference pattern is described by the equation

           d sin θ = m λ

In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle

         tan θ = y / L

as the angles are small,

         tan θ = sin θ / cos θ = sin θ

substituting

         sin θ = y / L

         d y / L = m λ

         y = m λ / d L

let's apply this formula for each wavelength

λ = 450 nm = 450 10⁻⁹ m

m = 5

d = 5.0 mm = 5.0 10⁻³ m

      y₁ = 5 450 10⁻⁹ / (5 10⁻³  1.4)

      y₁ = 3.21 10⁻⁴ m

we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m

      y₂ = 5 590 10⁻⁹ / (5 10⁻³  1.4)

      y₂=  4.21 10⁻⁴ m

the separation of these two lines is

        Δy = y₂ - y₁

        Δy = (4.21 - 3.21) 10⁻⁴ m

        Δy = 1 10⁻⁴ m

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current

Answers

Answer:

Explanation:

From the question we are told that

    The radius is  [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]

     The  current is  [tex]I = 4.5 \ A[/tex]

Generally the electric field is mathematically represented as

         [tex]E = \frac{J}{\sigma }[/tex]

Where [tex]\sigma[/tex] is the conductivity of  aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]

J is the current density which mathematically represented as  

      [tex]J = \frac{I}{A}[/tex]

Here A is the cross-sectional area which is mathematically represented as  

       [tex]A = \pi r^2[/tex]

       [tex]A = 3.142 * (1.4*10^{-3})^2[/tex]

       [tex]A = 6.158*10^{-6} \ m^2[/tex]

So

    [tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]

    [tex]J = 730757 A/m^2[/tex]

So

       [tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]

       [tex]E = 0.021 \ N/C[/tex]

A car travels at 45 km/h. If the driver breaks 0.65 seconds after seeing the traffic light turn yellow, how far will the car continue to travel before it begins to slow?

Answers

Answer:

8.1 m

Explanation:

Convert km/h to m/s.

45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s

Distance = speed × time

d = (12.5 m/s) (0.65 s)

d = 8.125 m

Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.

Answers

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

             

A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?

Answers

Answer:

Pls seeattached file

Explanation:

A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.

What is Resistance?

Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.

When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.

According to the question,

R = R₀ (1 + α ΔT)

(1 + 0.0135)R₀ = R₀(1 + α ΔT)

ΔT = (1 + 0.0135) / α

= 0.0135 / 0.0004

= 33.75 °C.

ΔT = [(1 - 0.0135) -1]/0.004

= -33.75 °C

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g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.

Answers

Answer:gamma ray

Explanation:

To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 2 cm

Answers

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

[tex]L = \frac{\mu_oN^2A}{l}[/tex]

where;

[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth

Answers

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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A 750 gram grinding wheel 25.0 cm in diameter is in the shape of a uniform solid disk. (we can ignore the small hole at the center). when it is in use, it turns at a consant 220 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 45.0 s with constant angular acceleration due to friction at the axle. What torque does friction exert while this wheel is slowing down?

Answers

Answer:

Torque = 0.012 N.m

Explanation:

We are given;

Mass of wheel;m = 750 g = 0.75 kg

Radius of wheel;r = 25 cm = 0.25 m

Final angular velocity; ω_f = 0

Initial angular velocity; ω_i = 220 rpm

Time taken;t = 45 seconds

Converting 220 rpm to rad/s we have;

220 × 2π/60 = 22π/3 rad/s

Equation of rotational motion is;

ω_f = ω_i + αt

Where α is angular acceleration

Making α the subject, we have;

α = (ω_f - ω_i)/t

α = (0 - 22π/3)/45

α = -0.512 rad/s²

The formula for the Moment of inertia is given as;

I = ½mr²

I = (1/2) × 0.75 × 0.25²

I = 0.0234375 kg.m²

Formula for torque is;

Torque = Iα

For α, we will take the absolute value as the negative sign denotes decrease in acceleration.

Thus;

Torque = 0.0234375 × 0.512

Torque = 0.012 N.m

The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.

Answers

Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.

Hope this helps!

Answer:

The electromagnetic waves reach Earth, while the mechanical waves do not

Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?

Answers

Answer:

L = 0.8 m

Explanation:

Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:

Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m

but,

Δx = λL/d

λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m

d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m

L = Distance between slits and screen = ?

Therefore, using the values, we get:

2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)

L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)

L = 0.8 m

A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.

Answers

Answer:

A) y = 3.56 mm

B) y = 3.56 mm

Explanation:

A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:

[tex] y = \frac{m\lambda L}{d} [/tex]

Where:

λ: is the wavelength = 546.1 nm

m: is first bright region = 1

L: is the distance between the screen and the plane of the parallel slits = 1.50 m

d: is the separation between the slits = 0.230 mm

[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]  

B) The distance between the first and second dark bands is:

[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]

Where:

[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]

[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]      

I hope it helps you!

Rank the following types of electromagnetic waves by the wavelength of the wave.

a. Microwaves
b. X-rays
c. Radio waves
d. Visible light

Answers

Explanation:

In order of Increasing Wavelength of the Electromagnetic Spectrum :

B) X rays

D) Visible light

A) Microwave

C) Radio Waves

Electromagnetic waves in order of decreasing wavelength  is X-rays,visible light,microwaves and radio waves.

What are electromagnetic waves?

The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.

The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.

In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.

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Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object

Answers

Answer:

Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.

Explanation:

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment

Answers

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

[tex]p_{f} = p_{i}[/tex]

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.

Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration

Answers

Answer:

The acceleration of the refrigerator together with the objects decreases.

Explanation:

If the mass of the refrigerator is increased by stacking more masses (objects) on it,

and the force applied remains constant, then we know from

F = ma

where

F is the applied force

m is the total mass of the refrigerator and the objects

a is the acceleration of the masses.

If F is constant, and m is increased, the acceleration will decrease

Answer:

The acceleration decreases.

Explanation:

its right

an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate

Answers

Answer:

The electric field  can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.

Explanation:

Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.

How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m

Answers

Answer:

Explanation:

volume of water being lifted

= π r² h , where r is radius of cylinder and h is height of cylinder

= 3.14 x5² x 10

= 785 m³

mass of water = 785 x 10³ kg

mass of this much of water is lifted so that its centre of mass is lifted by height

10 / 2 = 5m .

So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity

= 785 x 10³ x 9.8 x 5

= 38.465 x 10⁶ J  

What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 566 nm light? The index of refraction of the coating material is 1.46 and the index of the glass is 1.71.

Answers

Answer:

The  thickness is   [tex]t = 1.415 *10^{-7 } \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]

     The  index of refraction of glass is  [tex]n_g = 1.71[/tex]

     The index of refraction of the coating is  [tex]n= 1.46[/tex]

Generally the condition for destructive interference is  

         [tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]

Here m is the order of the interference pattern and given from the question that we are considering minimizing  reflection  m = 0

t = thickness of the coating

substituting values

         [tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]

    =>    [tex]t = 1.415 *10^{-7 } \ m[/tex]

           

A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?

Answers

Answer:

-0.24diopters

Explanation:

The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)

So 1/do +1/di = 1/f

1/infinity + 1/-4.2 = 1/f

1/f = 1/-4.2 = -0.24diopters

A simple arrangement by means of which e.m.f,s. are compared is known

Answers

Answer:

A simple arrangement by means of which e.m.f,s. are compared is known as?

(a)Voltmeter

(b)Potentiometer

(c)Ammeter

(d)None of the above

Explanation:

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.

Answers

Answer:

e. It is neither attracted nor repelled.

Explanation:

Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.

If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?

Answers

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3

Answers

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

     

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

           

            v₂ = 9 v

When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass

Answers

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.

Answers

Answer: Ф = 17.2657 ≈ 17°

Explanation:

we simply apply ET =0 about the ending of the rod

so In.g.L/2sinФ - In.a.L/2cosФ = 0

g.sinФ - a.cosФ = 0

g.sinФ = a.cosФ

∴ tanФ = a/g

Ф =  tan⁻¹ a / g

Ф = tan⁻¹ ( 10 / 32.17405)

Ф = tan⁻¹ 0.31080948777

Ф = 17.2657 ≈ 17°

Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°

3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.​

Answers

Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?

Answers

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as seen from the left)?

Answers

Answer:

The induced current is clockwise

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