Often in science it is helpful to talk about a representative example of the objects or phenomena being studied. However, you must always keep in mind that the average case is not always representative. For example, our Sun is often described as an "average" star in the Milky Way. In what sense is this statement true? In what sense is this statement seriously misleading? Do you think it is useful to characterize the stars in the Milky Way by simply citing our "average" Sun?

Answers

Answer 1

Explanation:

The statement "our sun is an 'average' sun" is  true when it is used to describe or characterize some unique physical properties of stars generally in the universe. 'Average' in this sense is used to define a typical sun such as, "stars should glow like our sun an average star."

The statement is used wrongly when used to in quantifying other stars in the universe, based on calculated values from our sun. In this case, we cannot truly say if our sun is a true representative average of other stars in the universe.

Yes! it is useful to characterize the milky way by simply citing the average sun. Properties like their ability to glow and radiate heat can be defined by citing an average star like our sun, so long as we don't translate it into citing quantitative properties of the sun as an average of our Milky Way Galaxy like the mass, temperature, etc.


Related Questions

Given that the mass of the Earth is 5.972 * 10^24 kg and the radius of the Earth is
6.371 * 10^6 m and the gravitational acceleration at the surface of the Earth is 9.81
m/s^2 what is the gravitational acceleration at the surface of an alien planet with
2.4 times the mass of the Earth and 1.9 times the radius of the Earth?
Although you do not necessarily need it the universal gravitational constant is G =
6.674 * 10^(-11) N*m^2/kg^2
9

Answers

Answer:

gₓ = 6.52 m/s²

Explanation:

The value of acceleration due to gravity on the surface of earth is given as:

g = GM/R²   -------------------- equation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

M = Mass of Earth

R = Radius of Earth

Now, for the alien planet:

gₓ = GMₓ/Rₓ²

where,

gₓ = acceleration due to gravity at the surface of alien planet

Mₓ = Mass of Alien Planet = 2.4 M

Rₓ = Radius of Alien Planet = 1.9 R

Therefore,

gₓ = G(2.4 M)/(1.9 R)²

gₓ = 0.66 GM/R²

using equation 1

gₓ = 0.66 g

gₓ = (0.66)(9.81 m/s²)

gₓ = 6.52 m/s²

A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.

Answers

the answer is c yw :D

g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is placed on the x axis at x = +30 cm. What is the magnitude of the total electrostatic force on a third particle (charge = -4.0 mC) placed at the origin (x = 0)? Group of answer choices

Answers

Answer:

Explanation:

We shall find electric field at origin due to two given charges sitting   on the either side of origin .

Total field will add up due to their same direction .

Field due to a charge Q

= 9 x 10⁹ x Q / R²  ;  R is distance of point , Q is charge

Field due to first charge

= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴

= 90 x 10¹⁰ N/C

Field due to second  charge

= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴

= 112.5 x 10¹⁰ N/C

Total field

= 202.5 x 10¹⁰ N/C

Force on given charge at origin

= charge x field

= 4 x 10⁻³ x 202.5 x 10¹⁰

= 810 x 10⁷ N .

A millionairess was told in 1992 that she had exactly 15 years to live. However, if she immediately takes off, travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:

Answers

Answer:

The expected year is 2017.

Explanation:

Total years that the millionaire to live = 15 years

Travel away from the earth at  = 0.8 c

This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:

[tex]T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years[/tex]

Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017

A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film

Answers

Answer:

3.867 μm

Explanation:

The index of refraction, μ = 1.6

Wavelength of the light, λ = 580 nm

N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have

(N2 - N1) λ = 2L (n2 - n1)

L = [(N2 - N1) * λ] / 2(n2 - n1)

L = (8 * 580) / 2(1.6 - 1.0)

L = 4640 nm / 1.2

L = 3867 nm or 3.867 μm

Therefore we can come to the conclusion that the thickness of the film is 3.867 nm

A 20 g bullet strikes a 1.5 kg wood block and becomes embedded in it (all horizontal motion). The coefficient of friction between the block and the surface is 0.3 and the impact makes the block move a distance of 7 m before it comes to rest. What was the muzzle speed (in m/s) of the bullet? Report your answer with two significant figures.

Answers

Answer:

Explanation:

Kinetic frictional force acting on the block = μ mg

where μ is coefficient of friction , m is mass of block.

.3 x 1.5 x 9.8 = 4.41 N .

Let v be the velocity of bullet + block after collision

kinetic energy of composite mass after the strike

= 1 /2 x 1.52 x v²

this will be equal to work done by friction .

.76 v² = 4.41 x 7

v² = 40.62

v = 6.37 m /s

Now we can obtain muzzle speed of bullet by applying conservation of momentum .

Let this speed be u

initial momentum of bullet

= .02 x u

final momentum of composite mass

= 1.52 x 6.37

.02 u = 1.52 x 6.37

u = 484.12 m /s .

= 480 m /s ( in two significant figures )

As light shines from air to another medium, i = 26.0 º. The light bends toward the normal and refracts at 32.0 º. What is the index of refraction? A. 1.06 B. 0.944 C. 0.827 D. 1.21

Answers

Explanation:

It is given that,

Angle of incidence from air to another medium, i = 26°

The angle of reflection, r = 32°

We need to find the refractive index of the medium. The ratio of sine of angle of incidence to the sine of angle of reflection is called refractive index. It can be given by :

[tex]n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (26)}{\sin (32)}\\\\n=0.82[/tex]

So, the index of refraction is 0.82. Hence, the correct option is C.

A computer has a mass of 3 kg. What is the weight of the computer?
A. 288 N.
B. 77.2 N
C. 3N
D. 29.4 N

Answers

Answer:

29.4 N

Option D is the correct option.

Explanation:

Given,

Mass ( m ) = 3 kg

Acceleration due to gravity ( g ) = 9.8 m/s²

Weight ( w ) = ?

Now, let's find the weight :

[tex]w \: = \: m \times g[/tex]

plug the values

[tex] = 3 \times 9.8[/tex]

Multiply the numbers

[tex] = 29.4 \: [/tex] Newton

Hope this helps!!

best regards!!

An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by rho(r)=αr,where α is a positive constant.
A). What is the magnitude of the electric field at a distance r from the center of the shell, where a Express your answer in terms of the variables α, a, r, and electric constant ϵ0.
B) .A point charge
q is placed at the center of the hollow space, at r=0. What value must q have (sign and magnitude) in order for the electric field to be constant in the region a Express your answer in terms of the variables α, a, and appropriate constants.
C). What then is the value of the constant field in this region?
Express your answer in terms of the variable αand electric constant ϵ0.

Answers

Answer:

E   = α/2∈₀ [ 1 - a²/r² ]

Ф = α/2∈₀

Explanation:

Using Gauss Law:

    ρ(r) = a/r, dA

          = 4 π r²d r

    Ф = [tex]\int\limits^r_a[/tex] ρ(r')dA

    Ф[tex]_{encl}[/tex] = [tex]\int\limits^r_a[/tex] ρ(r')dA

             = 4πα [tex]\int\limits^r_a[/tex] r'dr'

Ф[tex]_{encl}[/tex]     = 4 π α 1/2(r²-a²)

E(4πr²) = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀

           = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀(4πr²)

           = α (r² - a²) / 2 ∈₀ (r²)

           = α/2∈₀ [ r²/r² - a²/r² ]

      E   = α/2∈₀ [ 1 - a²/r² ]

Electric field of the point charge:

E[tex]_{q}[/tex] = q / 4π∈₀r²

[tex]E_{total}[/tex] = α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²

For [tex]E_{total}[/tex]  to be constant:

- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²

-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀

= α - αa² + αa² / 2 ∈₀

= α /2 ∈₀

Hence:

Ф = α/2∈₀

When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the copper is cooled with liquid nitrogen, the resistivity of the copper drops dramatically. How will this affect the speed at which the magnet falls toward the copper

Answers

Answer:

The speed at which the magnet falls through the copper block will be reduced dramatically.

Explanation:

Eddy's current are loops of electrical current induced within conductors by a changing magnetic field in the conductor. Eddy's current is proportional to the the magnetic field strength, the rate of change of flux, the area of the loop, and is inversely proportional to the resistivity of the material. Eddy currents flows perpendicularly to the magnetic field, and in closed loops within conductors.

Reducing the resistivity of the copper will increase the Eddy current on the copper, which will in turn increase the opposition to the action producing the flux change (the falling magnet through the copper block). The result is that the speed at which the magnet falls through the copper block will be reduced dramatically.

If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the food, how should the person aim the arrow containing the food?

Answers

Answer:

The options are

A. He should aim it at the monkey

B. He should aim it below the monkey

C. He should aim it above the monkey

D. None of the above

The answer is A. He should aim it at the monkey

This is because the monkey has a large surface area and a bigger body mass. This will make aiming the food at the monkey feasible in it getting it as it could use other parts of the body to get the food aimed at it. The monkey won’t reach the food when falling if it is aimed above it. It also won’t get to the it when it is shot at below it.

Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?

Answers

Answer:

0.11m

Explanation:

let's assume the boat is of uniform construction

Ignoring friction losses

Also assume the origin is at the end of the boat originally with the heavier person

the center of mass of the whole system will not change relative to the water when the two swap ends

Originally, the center of mass is

85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin

after the swap, the center of mass is

50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin

The center of mass has shifted

1.14-1.030 = 0.11m

as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat

The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at speed c toward the space vehicle. What is the speed of the light signal as viewed from the space vehicle

Answers

Answer:

The speed of the light signal as viewed from the observer is c.

Explanation:

Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.

Force and distance are used to calculate work. Work is measured in which unit? joules watts newtons meters

Answers

Answer:

The unit of work is joules

Force and displacement are used to calculate the work done by an object. This work is measured in the units of Joules. Thus, the correct option is A.

What is Work?

Work can be defined as the force that is applied on an object which shows some displacement. Examples of work done include lifting an object against the Earth's gravitational force, and driving a car up on a hill. Work is a form of energy. It is a vector quantity as it has both the direction as well as the magnitude. The standard unit of work done is the joule (J). This unit is equivalent to a newton-meter (N·m).

The nature of work done by an object can be categorized into three different classes. These classes are positive work, negative work and zero work. The nature of work done depends on the angle between the force and displacement of the object. Positive work is done if the applied force displaces the object in its direction, then the work done is known as positive work. Negative work is opposite of positive work as in this work, the applied force and displacement of the object are in opposite directions to each other and zero work is done when there is no displacement.

Therefore, the correct option is A.

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A bug is sitting on the edge of a rotating disk. At what angular velocity will the bug slide off the disk if its radius is 0.241 m, the coefficient of static friction between the bug and disk is 0.321, and the coefficient of kinetic friction is 0.102

Answers

Answer:

ω = 3.61 rad/sec

Explanation:

Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.

μmg = mv^2/r = mω^2r

Thus;

μg = ω^2r

ω^2 = μg/r

ω = √(μg/r)

ω = √(0.321 * 9.8)/0.241

ω = √(13.05)

= 3.61 rad/sec

gravity can be described as..?

A. an magnetic force found in nature

B.the force that moves electrical charges

C.the force that repels object with like chargers

D.the force of attraction between two objects

Answers

Answer:

D

Explanation:

Gravity is the force of attraction between two objects.

Each object creates a gravitational field in wich every other object is affected by it.

a figures skater rotating at 5 rads with arms extended has a moment of inertia of 2.25 kg. if the arms are pulled in so the moment of inertia decrease to 1.8 what is the final angular speed

Answers

Answer:

The final angular speed is 6.25 rad/s

Explanation:

Given;

initial angular speed, ω₁ = 5 rad/s

initial moment of inertia, I₁ = 2.25 kg.m²

Final moment of inertia, I₂ = 1.8 kg.m²

final angular speed, ω₂ = ?

Based on conservation of angular momentum, we will have the following expression;

ω₁I₁ = ω₂I₂

ω₂ = (ω₁I₁ ) / I₂

ω₂ = (5 x 2.25) / 1.8

ω₂ = 6.25 rad/s

Therefore, the final angular speed is 6.25 rad/s

A 285-kg object and a 585-kg object are separated by 4.30 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object placed midway between them.

Answers

Answer:

The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

Explanation:

Given;

first object with mass, m₁ = 285 kg

second object with mass, m₂ = 585 kg

distance between the two objects, r = 4.3 m

The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m

Gravitational force between the first object and the 42 kg object;

[tex]F = \frac{GMm}{r^2}[/tex]

where;

G = 6.67 x 10⁻¹¹ Nm²kg⁻²

[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]

Gravitational force between the second object and the 42 kg object

[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]

Magnitude of net gravitational force exerted on 42kg object;

F = 3.545x 10⁻⁷ N  -  1.727 x 10⁻⁷ N

F = 1.818 x 10⁻⁷ N

Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

have an electrical charge of +1, while
have an electrical charge of -1.
A. Neutrons, electrons
B. Protons, electrons
C. Electrons, neutrons
D. Electrons, protons

Answers

Answer:

B

Explanation:

Protons have a positive electrical charge of +1,

Electrons have a negative charge of -1,

Neutrons have a neutral charge of about 0.

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m 3 kg/m3 , length 81.2 cmcm and diameter 2.60 cmcm from a storage room to a machinist. Calculate the weight of the rod, www. Assume the free-fall acceleration is ggg = 9.80 m/s2m/s2 .

Answers

Answer:

The weight of the rod is 32.87 N

Explanation:

Density of the rod = 7800 kg/m

length of the rod = 81.2 cm = 0.812 m

diameter of rod = 2.60 cm = 0.026 m

acceleration due to gravity = 9.80 m/s^2

The rod can be assumed to be a cylinder.

The volume of the rod can be calculated as that of a cylinder, and can be gotten as

V = [tex]\frac{\pi d^{2} l}{4}[/tex]

where d is the diameter of the rod

l is the length of the rod

V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3

We know that the mass of a substance is the density times the volume i.e

mass m = ρV

where ρ is the density of the rod

V is the volume of the rod

m = 4.3 x 10^-4 x 7800 = 3.354 kg

The weight of a substance is the mass times the acceleration due to gravity

W = mg

where g is the acceleration due to gravity g = 9.80 m/s^2

The weight of the rod W = 3.354 x 9.80 = 32.87 N

A 43.0-g toy car is released from rest on a frictionless track with a vertical loop of radius R. The initial height of the car is h = 4.05R.

Required:
a. What is the speed of the car at the top of the vertical loop?
b. What is the magnitude of the normal force acting on the car at the top of the vertical loop?

Answers

Answer:

A.) 909 cm/s

B.) 33075 N

Explanation:

A.) Given that the

Mass M = 43 g

Height h = 4.05 R

Radius r = R

At the top of the loop, the maximum potential energy P.E = mgh

Substitutes m and h into the formula where g = 9.8 m/s^2 = 9610.517 cm/s^2

P.E = 43 × 9610.517 × 4.05R

P.E = 1673671.536R J

According to conservative of energy

The maximum P.E = maximum K.E

But K.E = 1/2mv^2

1673671.536R = 1/2mv^2

Substitutes for mass m into the formula

1673671.536R = 1/2× 4.05R × v^2

The R will cancel out

Cross multiply

4.05 v^2 = 3347343.072

V^2 = 3347343.072 / 4.05

V^2 = 826504.4622

V = sqrt( 826504.4622)

V = 909 cm/s

B.) At the top of the loop, the centripetal force = the sum of the normal force N and the weight W of the car. That is,

MV^2/R = N + W

Make N the subject of formula

N = mv^2/ R - W

Where W = mg

Substitute all the parameters into the formula

N = (4.05R × 909^2) /R - 4.05R × 9610.517

N = 3346438.05 - 38922.59

N = 3307515 N

a bear has a mass of 500kg and 100,000 J of mechanical kinetic energy. What is the speed of the bear? (velocity) Can someone please answer with the formula included? Step by step pls.

Answers

Answer:

Velocity of bear (v) = 20 m/s

Explanation:

Given:

Mass of bear (m) = 500 kg

Mechanical kinetic energy (K.E) = 100,000 J

Find:

Velocity of bear (v) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2(m)(v)²

100,000 = 1/2(500)(v)²

200,000 = 500(v)²

400 = (v)²

Velocity of bear (v) = 20 m/s

The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, what condition would he have?
A. Nearsightedness
B. Farsightedness
C. Neither nearsightedness nor farsightedness

Answers

Answer:

A. Nearsightedness

Explanation:

A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.

At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.

Two charges, +9 µC and +16 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force (in N) on a −7 nC charge when placed at the following locations.
(a) halfway between the two
(b) half a meter to the left of the +9 µC charge
(c) half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.)

Answers

Answer:

A) 1.76U×10⁻³N

B) 2.716×10⁻³N

C) 264.5⁰

Explanation:

See detailed workings for (a), (b), (c) attached.

Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short wavelength. It has a long wavelength. It has a high intensity. It has a low intensity.

Answers

Answer:

Longest wavelength, lowest intensity

Explanation:

Answer:

It has a long wavelength

Explanation:

GRADPOINT

A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N

Answers

The velocity and force are required.

The speed of the racket is 8.7 m/s

The required force is 471.43 N.

[tex]m_1[/tex] = Mass of racket = 1000 g

[tex]m_2[/tex] = Mass of ball = 60 g

[tex]u_1[/tex] = Initial velocity of racket = 12 m/s

[tex]u_2[/tex] = Initial velocity of ball = -15 m/s

[tex]v_1[/tex] = Final velocity of racket

[tex]v_2[/tex] = Final velocity of ball = 40 m/s

[tex]\Delta t[/tex] = Time = 7 ms

The equation of the momentum will be

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]

Force is given by

[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]

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The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?

Answers

Answer:

1.03A

Explanation:

For computing the magnitude of the current in the circuit we need to do the following calculations

LCR circuit impedance

[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]

= 148.7Ω

Now the phase angle is

[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]

Now the rms current flowing in the circuit is

[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]

= 0.98 A

The current flowing in the circuit is

[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]

= 1.39 A

And, finally, the current across the generator is

[tex]I'= I cos \phi[/tex]

[tex]= (1.39) cos 42.3^{\circ}[/tex]

= 1.03A

Hence, the magnitude of the circuit current is 1.03A

What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.

Answers

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  wavelength is  [tex]\lambda = 622 nm[/tex]

Explanation:

  From the question we are told that

    The distance of the slit to the screen is  [tex]D = 5 \ m[/tex]

    The order of the fringe is m  =  6

     The distance between the slit is  [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]

    The fringe distance is  [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]

Generally the for a dark fringe the fringe distance is  mathematically represented as

        [tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]

=>     [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]

substituting values

=>      [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]

=>     [tex]\lambda = 6.22 *10^{-7} \ m[/tex]

       [tex]\lambda = 622 nm[/tex]

The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.

Answers

Answer:

constant horizontal force developed in the coupling C = 11.25KN

the friction force developed between the tires of the truck and the road during this time is 33.75KN

Explanation:

See attached file

The friction force between the tires of the truck and the road is 22500 N.

Calculating the friction force:

It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.

Distance traveled before coming to rest, s = 10m

The final velocity of the truck will be zero, v = 0

When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.

The frictional force is given by:

f = -ma

the acceleration of the truck = -a

The negative sign indicates that the acceleration is opposite to the motion.

Applying the third equation of motion we get:

v² = u² -2as

0 = 15² - 2×a×10

225 = 20a

a = 11.25 m/s²

So the magnitude of frictional force is:

f = ma = 2000 × 11.25 N

f = 22500 N

Learn more about friction force:

https://brainly.com/question/1714663?referrer=searchResults

The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.

Answers

Answer:

27°

Explanation:

The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)

So theta = arcsin(0.45)

=27°

The angle between the wire and the magnetic field is 27°.

Calculation of the angle:

Since The magnetic force per meter on a wire is measured to be only 45 %

So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field

Therefore,

theta = arcsin(0.45)

=27°

Hence, The angle between the wire and the magnetic field is 27°.

Learn more about wire here: https://brainly.com/question/24733137

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