A very fine thread is placed between two glass plates on one side and the other side is touching to form a wedge. A beam of monochromatic light of wavelength 600 nm illuminates the wedge and 178 bright fringes are observed. What is the thickness of the thread?

Answer:

the total charge on the disk 256pi Coulombs

Explanation:

Pls see attached file

Answer:

The mass of the object is 745000 units of the sun

Explanation:

We know that the centripetal force with which the stars orbit the object is represented as

[tex]F_{c}[/tex] = [tex]\frac{mv^{2} }{r}[/tex]

and this centripetal force is also proportional to

[tex]F_{c}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]

where

m is the mass of the stars

M is the mass of the object

v is the velocity of the stars = 10^6 m/s

r is the distance between the stars and the object = 10^14 m

k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

We can equate the two centripetal force equations to give

[tex]\frac{mv^{2} }{r}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]

which reduces to

[tex]v^{2}[/tex] = [tex]\frac{kM}{r}[/tex]

and then finally

M = [tex]\frac{rv^{2} }{k}[/tex]

substituting values, we have

M = [tex]\frac{10^{14}*(10^{6})^{2} }{6.67*10^{-11} }[/tex] = 1.49 x 10^36 kg

If the mass of the sun is 2 x 10^30 kg

then, the mass of the the object in units of the mass of the sun is

==> (1.49 x 10^36)/(2 x 10^30) = 745000 units of sun

Answer:

The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.

Explanation:

Answer:

The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.

Explanation:

Edmentum sample answer

Answer:

Heat Flow Rate : ( About ) 87 W

Explanation:

The heat flowing out of the system each minute, will be represented by the following equation,

Q( cup ) + Q( water ) = m( cup ) [tex]*[/tex] c( al ) [tex]*[/tex] ΔT + m( w ) [tex]*[/tex] c( w ) [tex]*[/tex] ΔT

So as you can see, the mass of the aluminum cup is 150 grams. For convenience, let us convert that into kilograms,

150 grams = .15 kilograms - respectively let us convert the mass of water to kilograms,

800 grams = .8 kilograms

Now remember that the specific heat of aluminum is 900 J / kg [tex]*[/tex] K, and the specific heat of water = 4186 J / kg [tex]*[/tex] K. Therefore let us solve for " the heat flowing out of the system per minute, "

Q( cup ) + Q( water ) = .15 [tex]*[/tex] ( 900 J / kg [tex]*[/tex] K )  [tex]*[/tex] 1.5 + .8 [tex]*[/tex] ( 4186 J / kg [tex]*[/tex] K ) [tex]*[/tex] 1.5,

Q( cup ) + Q( water ) = 5225.7 Joules

And the heat flow rate should be Joules per minute,

5225.7 Joules / 60 seconds = ( About ) 87 W

Answer:

Explanation:

coefficient of linear expansion α = 2.9 x 10⁻⁵

coefficient of volume expansion γ = 3 x 2.9 x 10⁻⁵ = 8.7 x 10⁻⁵

[tex]d_t = d_0( 1 - \gamma t )[/tex]

[tex]d_{82} = 1.13\times 10^4( 1 - 8.7\times 10^{-5}\times82 )[/tex]

= 1.13 x 10⁴ - 806.14 x 10⁻¹

= 1.13 x 10⁴ - 0.00806 x 10⁴

= 1.1219 x 10⁴ kg / m³

b ) mass of the sample will remain the same as mass does not increase or decrease with temperature .

Answer:

D. 0.1

Explanation:

Using transformer equation,

N2/N1 = I1/I2................... Equation 1

Where N2 = secondary coil, N1 = primary coil, I1 = input current, I2 = output current.

make I2  the subject of the equation

I2 = I1/(N2/N1)............ Equation 2

From equation 2 above, For the output current of the secondary coil to be 10 times the input current, N2/N1 = 0.1

Hence the right option is D. 0.1

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

Hello. This question is incomplete. The full question is:

A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity. (There may be more than one correct choice.)

A) The potential is lowest, but not zero, at the center of the sphere.  B) The potential at the center of the sphere is zero.  C) The potential at the center of the sphere is the same as the potential at the surface.  D) The potential at the surface is higher than the potential at the center.  E) The potential at the center is the same as the potential at infinity

Answer:

C) The potential at the center of the sphere is the same as the potential at the surface.

Explanation:

When a conductive sphere has charges that distribute evenly on its surface, it means that its interior has a zero charge cap. As a result, the outside of this sphere has a charge distribution that will be the same if the center of the sphere were charged. In this way, the center and the surface of the sphere become identical in relation to the point charge potential. In other words, this means that the null interior of the sphere has a constant potential that makes the distribution of charges within the sphere exactly equal to the distribution of charges outside the sphere.

The statement that should be true regarding the sphere is The potential at the center of the sphere.

Here the normal formula should be used i.e.  kq/R that should be determined by considering the potential at infinity also it does not contain any intervening dielectric like zero.

Also,

kq/R+c,

Here c is constant is necessary to fit with respect to the zero potential.

Hence, The statement that should be true regarding the sphere is The potential at the center of the sphere.

Learn more about sphere here: https://brainly.com/question/16684376

Answer:

The  force on the foundation wall is   [tex]F_f = 191394 \ N[/tex]

Explanation:

From the question we are told that

     The  depth of the hole's vertical side is  [tex]d = 2.10 \ m[/tex]

      The  width of the hole is  [tex]b = 8.90 \ m[/tex]

      The distance of the concrete wall from the front of the cellar is  [tex]c = 0.189 \ m[/tex]

Generally the area which the water from the drainage covers is mathematically represented as

        [tex]A = d * b[/tex]

substituting values

        [tex]A = 2.10 * 8.90[/tex]

       [tex]A = 18.69 \ m^2[/tex]

Now the gauge pressure exerted on the foundation wall is mathematically evaluated as

          [tex]P_g = \rho * d_{avg} * g[/tex]  

Here  is the average height foundation wall where the pressure of the water is felt and it is evaluated as

         [tex]d_{avg} = \frac{h_1 + h_2 }{2}[/tex]

where [tex]h_1[/tex] at the height at bottom of the hole which is equal to  [tex]h_1 = 0[/tex]

and  [tex]h_2[/tex] is the height at the top of the hole [tex]h_2 = d = 2.10[/tex]

        [tex]d_{avg} = \frac{0 + 2.10 }{2}[/tex]

       [tex]d_{avg} = 1.05[/tex]

Where  [tex]\rho[/tex] is the density of water with constant value [tex]\rho = 1000 \ kg/m^3[/tex]

substituting values

          [tex]P_g = 1000 * 1.05 * 9.8[/tex]

         [tex]P_g = 10290 \ Pa[/tex]

Then the force exerted by the water on the foundation wall mathematically represented as

      [tex]F_f = P_g * A[/tex]

substituting values

      [tex]F_f = 10290 * 18.69[/tex]

       [tex]F_f = 191394 \ N[/tex]

Answer:

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Answer:

32

Explanation:

Explanation:

It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.

We need to find the speed before and after the ball bounce.

Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,

[tex]\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s[/tex]

Let v is the final speed when it bounces to a height of 1.5 m. So,

[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s[/tex]

So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.

Answer:

9*10^13 J

Explanation:

Given that

mass of the material, m = 0.001 kg

speed of light, c = 3*10^8 m/s

To solve this, we would be adopting Einstein's mass - energy relation...

E = mc²

Where

E = Energy, in joules

m = mass, in kg

c = speed of light, in m/s

E = 0.001 * (3*10^8)²

E = 0.001 * 9*10^16

E = 9*10^13 J

Thus, the total energy released would be 9*10^13 J

Answer:

a) 6738.27 J

b) 61.908 J

c)  [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

[tex]I[/tex] =  [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] =  [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

[tex](I_{1} +I_{2} )w[/tex]

where the subscripts 1 and 2 indicates the values first and second  flywheels

[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

where m is the mass of the car

[tex]v_{car}[/tex] is the velocity of the car

Equating the energy

2246.09 =  [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

making m the subject of the formula

mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Answer:

The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.

Answer:

The merry-go-round's angular velocity 23.84 RPM

Explanation:

Given;

diameter of merry go round, d = 3 m

radius of the merry go round, R = 1.5 m

mass of the merry go round, m = 300 kg

angular velocity = 23 rpm

velocity of John, v = 4.4 m/s

mass of John, m = 30 kg

Apply conservation of angular momentum;

[tex]L_i = L_f[/tex]

[tex]I \omega_i + mvR = (I + mR^2)\omega _f[/tex]

where;

I is moment of inertia of disk

[tex]I = \frac{1}{2} mR^2\\\\I = \frac{1}{2} *300*1.5^2\\\\I = 337.5 \ kg.m^2[/tex]

Substitute in this value in the above equation;

[tex]337.5(2\pi \frac{23}{60} ) + (30*4.4*1.5) = (337.5 + 30*1.5^2) \omega_f\\\\812.9925 \ + \ 198 = 405 \omega _f\\\\1010.9925 = 405 \omega _f\\\\\omega _f = \frac{1010.9925}{405} \\\\\omega _f = 2.496 \ rad/s[/tex]

1 rad/s = 9.5493 rpm

2.496 rad/s = 23.84 RPM

Therefore, the merry-go-round's angular velocity 23.84 RPM

Answer:

There are 586maxima

Explanation:

Pls see attached file

Answer:

(a) 1.5×10⁻¹¹ m.

(b) 6.7×10⁻² m

Explanation:

Note: All Electromagnetic wave travels in with the same speed, which 3×10⁸ m/s

(a) Give a frequency of 2.00×10¹⁹ Hz.

Using the equation of a wave,

V = λf................ Equation 1

Where V = Speed of electromagnetic wave, λ = wavelength, f = frequency.

make λ the subject of the equation

λ = V/f................. Equation 2

Given: f = 2.00×10¹⁹ Hz.

Constant: v = 3×10⁸ m/s.

Substitute into equation 2

λ = 3×10⁸/2.00×10¹⁹

λ = 1.5×10⁻¹¹ m.

(b) Similarly using

λ = v/f

Given: f = 4.5×10⁹ Hz, and v = 3×10⁸ m/s.

Substitute these values into equation 2 above.

λ = 3×10⁸/4.5×10⁹

λ = 6.7×10⁻² m

Complete  Question

In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 57.9 cm from the slits?

Answer:

a

  [tex]\theta = 0.3819^o[/tex]

b

  [tex]y = 0.00386 \ m[/tex]

Explanation:

From the question we are told that

    The slit separation is  [tex]d = 150 \lambda[/tex]

    The  distance from the screen is  [tex]D = 57.9 \ cm = 0.579 \ m[/tex]

Generally the condition for constructive interference is mathematically represented as

            [tex]dsin (\theta ) = n * \lambda[/tex]

=>        [tex]\theta = sin ^{-1} [\frac{n * \lambda }{ d } ][/tex]

where  n is the order of the maxima  and value is 1 because we are considering the central maximum and an adjacent maximum

     and  [tex]\lambda[/tex] is the wavelength of the light

So

       [tex]\theta = sin ^{-1} [\frac{ 1 * \lambda }{ 150 \lambda } ][/tex]

       [tex]\theta = 0.3819^o[/tex]

Generally the distance between the maxima is mathematically represented as

       [tex]y = D tan (\theta )[/tex]

=>    [tex]y = 0.579 tan (0.3819 )[/tex]

=>    [tex]y = 0.00386 \ m[/tex]

Answer:

The apple fell at a distance of 4.17 m.

Explanation:

Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.

When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.

In the International System of Units, work is measured in Joule. Joule is equivalent to Newton per meter.  

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.  

Work=Force*distance* cosine(angle)

On the other hand, Newton's second law says that the acceleration of a body is proportional to the resultant of forces on it acting and inversely proportional to its mass. This is represented by:

F=m*a

where F is Force [N], m is Mass [kg] and a Acceleration [m / s²]

In this case, the acceleration corresponds to the acceleration of gravity, whose value is 9.81 m / s². So you have:

Replacing:

9 J= 2.1582 N* distante* 1

Solving:

[tex]distance=\frac{9J}{2.1582 N*1}[/tex]

distance= 4.17 m

The apple fell at a distance of 4.17 m.

Answer:

3859 g

Explanation:

1 pound = 0.454 kg

therefore, 8.50 ponds = 0.454*8.50 = 3.859

to covert kilograms into grams you need to multiply it by 1000

=3.859*1000

= 3859 grams

Answer:

[tex]500\text{N} (490\text{N}) (490.5\text{N})[/tex]

Explanation:

The reaction force is the force that is in the perpendicular direction to the wall.

We have an angle and a hypotenuse, we need to find the adjacent angle - so we can just use cos:

[tex]cos(\theta)=\frac{\text{adj}}{\text{hyp}}\\\text{hyp}*cos(\theta)=\text{adj}\\100*cos(60)=100*0.5=50\text{kg}[/tex]

However, we would like a force and not a mass.

[tex]W=mg\\W=50g\\W=500\text{N} (490\text{N}) (490.5\text{N})[/tex]

Answer 1 if you use g as 10, answer 2 if you're studying mechanics in maths, answer 3 if you're studying mechanics in physics.

Answer:

-Some synthetic polymers use materials from Earth that are nonrenewable.

-They can end up as waste products that sometimes can’t be recycled.

-They sometimes release toxins into the environment.

Explanation:

Synthetic polymers are types of polymers that are man made, that is they are polymers that are made in the labs or industries from chemical substances.

Such polymers include nylon, polyester, and polyethylene among others. These polymers are used to make many clothes and plastics that we use and see around.

Synthetic polymers have a range of disadvantages which includes. being non-biodegradable, these polymers and especially plastics may end up as waste products and pollutes the environment and end up in livers and lakes and this would be toxic for aquatic animals.

Answer:

2.39 Ω

Explanation:

Given that

Number of winnings on the coil, = 250 turns

Radius if the copper wire, r(c) = 1.3/2 = 0.65 mm

Radius of single cylinder layer, R = 12 cm

Length of the cylinderical coil, L = 250 * 2π * 12 = 188.4 m

Resistivity of copper, ρ = 1.69*10^-8 Ωm

Area is πr(c)², which is

A = 3.142 * (0.65*10^-3)²

A = 3.142 * 4.225*10^-7

A = 1.33*10^-6 m²

The formula for resistance is given as

R = ρ.L/A, if we substitute, we have

R = (1.69*10^-8 * 188.4) / 1.33*10^-6

R = 3.18*10^-6 / 1.33*10^-6

R = 2.39 Ω.

Therefore, the resistance is 2.39 Ω

Answer:

The maximum speed of the motorcycle should be 21 m/s

Explanation:

Since the hill is considered to be a circular arc, the motorcycle will experience centripetal force that tends to flip it away from the center of the hill.

Since the motorcycle does not lose contact with the ground, it means that the weight of the motorcycle downwards just balances the centripetal force on the motorcycle.

we know that the centripetal force on the motorcycle is equal to

centripetal force = [tex]\frac{mv^{2} }{r}[/tex]

where m is the mass of the motorcycle,

v is the velocity of the motorcycle,

and r is the radius of the hill = 45.0 m

Also we now that the weight of the motorcycle is equal to

weight = mg

where m is still the mass of the motorcycle,

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

Equating the both forces since they are equal, we'll have

[tex]\frac{mv^{2} }{r}[/tex] = mg

the mass of the motorcycle will cancel out, and we'll be left with

[tex]v^{2} = gr[/tex]

[tex]v = \sqrt{gr}[/tex]

[tex]v = \sqrt{9.81*45}[/tex]

[tex]v = \sqrt{441.45}[/tex]

[tex]v[/tex] = 21 m/s

Answer:

A: 28°

B. 1x10^-3M

Explanation:

See attached file

Answer:

Your answer is( D) - Arago

Answer:

Explanation:

Magnetic field at a a point R distance away

B = μ₀ / 4π X 2I / R where I is current

Magnetic field due to one current

=  10⁻⁷ x 2 x 24 / 1 x 10⁻³

48 x 10⁻⁴ T

Magnetic field due to other current

=  10⁻⁷ x 2 x 24 / 3x 10⁻³

16 x 10⁻⁴ T

Total magnetic field , as they act in opposite direction, is

= (48 - 16 ) x 10⁻⁴

32 x 10⁻⁴ T .

Answer:

a)   d sin θ = m λ₀ / n

b)   d sin θ = (m + ½) λ₀ / n

c)    d = 2,439 10⁻⁷ m

Explanation:

For the interference these rays of light we must take as for some aspects,

* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2

* when the ray penetrates the lens the donut length changes by the refractive index

            λ = λ₀ / n

now let's write the destructive interference equation for these lightning bolts

           d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n

           d sin θ = m λ₀ / n

b) now nc> np

in this case there is no phase change in the reflected ray and the equation for destructive interference remains

             d sin θ = (m + ½) λ₀ / n

c) select the value of nc = 2.05 of the ZnO

we calculate the thickness of the film (d)

            d = m λ / (n sin 90)

in this type of interference the observation is normal, that is, the angle is 90º)

           d = 1 500 10-9 / (2.05 1)

           d = 2,439 10⁻⁷ m

d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference