Experiment to find ways to make rainbows.
a) Insert at least one setup where light passing through a prism gives a rainbow and describe why a rainbow is formed.
b) Explain why only some types of light will yield rainbows.

Answers

Answer 1

Answer:

Explanation:

a) To get a rainbow from a prism arrangement, we will need

A triangular prismA black cardboard boxA source of white light (light from the window will suffice)A pocket knife

First, you cut a slit in one end of the cardboard with the pen knife.

Next you open up a space on top of the cardboard through which you can observe the experiment and its result.

Next, you place the triangular prism with its slant face facing the the cut slit.

Finally, position the slit to face the light from the open window, and adjust the prism till the projected bands of colored light (rainbow) is very much obvious on the other end of the box, opposite the slit.

b) For a light to yield rainbow, it most be composed of different component colors of light. The colors of light is due to the difference in wavelength, and dispersion is due to the different in the wavelengths of the component light. So to get rainbow from a light source, the light must not be monochromatic. This means that only light composed of component light of different colors can produce rainbow. Light from the sun for example is composed of 7 distinct colors of light, and white light can be created with just three colors; blue, green, and red light.


Related Questions

An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
What is the average velocity of the airplane? Round your answer to the nearest whole number.

Answers

Maybe it is around 300

Answer:

433

Explanation:

(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.


pF

(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV

Answers

Explanation:

(a) Given that,

Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]

The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]

The dielectric constant of, k = 2.1

When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :

[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]

Putting all the values we get :

[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]

(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]

The voltage difference between the plates at this critical voltage is given by :

[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]

or

V = 0.6 kV

We have that the Capacitance and potential difference is mathematically given as

[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF



Capacitance &potential difference

Question Parameters:

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.

a)

Generally the equation for the Capacitance  is mathematically given as

[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]

C=334.68pF

b)

Generally the equation for the Capacitance  is mathematically given as

[tex]Vmax=\frac{Q}{C}[/tex]

Where

Q is the charge on the plates, and hence not given

Therefore, maximum potential difference is

[tex]Vmax=\frac{Q}{334.68pF}[/tex]

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A particle moves along the x axis. In order to calculate the torque on the particle, you need to know:

a. the rotational inertia of the particle
b. the velocity of the particle
c. the mass of the particle
d. the kinetic energy of the particle
e. the point about which the torque is to be calculated

Answers

Answer:

e. the point about which the torque is to be calculated

Explanation:

torque is the product of a force and a distance

the point about which the torque is calculated is required to know the distance.

None of the other terms are relevant as they refer to mass or its equivalent, and velocity. Force is not mentioned in any of them.

A particle moves along the x-axis. In order to calculate the torque on the particle, you need to know the point about which the torque is to be calculated. Therefore, option E is correct.

What is torque ?

The rotating equivalent of force is torque. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.

Ancient Romans gave these necklaces the term "torque" by describing them as twisted and spiral screw-shaped using the Latin word "torquere," which also means "twisting" and "turning."

It's critical to realize that torque, which has to do with your motor's power in terms of rotational force, is not the same thing as speed. Find a motor with a top speed if you require more motor speed, and a motor with a motor torque that is maximized if you need more rotational force.

Thus, option E is correct.

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Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.
A. Drawn from the top of the object so that it passes through the center of the lens at the optical axis.
B. Drawn from the top of the object so that it passes through the focal point on the same side of the lens as the object.
C. Drawn parallel to the optical axis from the top of the object.
D. Ray bends parallel to the optical axis.
E. Ray bends so that it passes through the focal point on the opposite side of the lens as the object.
F. Ray does not bend.

Answers

Answer:

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Explanation:

In this exercise you are asked to relate each with the answers

In general, in the optics diagram,

* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point

* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate

* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.

With these statements, let's review the answers

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

g A certain elevator cab has a total run of 195 m and a maximum speed is 306 m/min, and it accelerates from rest and then back to rest at 1.19 m/s2. (a) How far does the cab move while accelerating to full speed from rest

Answers

Answer:

About 23 meters

Explanation:

To do this, you'll want to apply one of the kinematic equations to find the time it takes for the cabin to reach max velocity from rest. (Use the max velocity as V_f and V_i=0)

Then, you can find the distance travelled during the acceleration by equating the acceleration to the change in distance of the time squared.

My work is in the attachment, comment if you have any questions.

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.

Required:
How far away did the impact occur?

Answers

Answer:

The distance is [tex]d = 193.6 \ m[/tex]

Explanation:

From the question we are told that

   The time interval between the sounds is  k[tex]t_1 = k + t_2[/tex] =  0.50 s

    The  speed of sound in air is  [tex]v_s = 343 \ m/s[/tex]

    The  speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]

 

Generally the distance where the collision occurred is  mathematically represented as

          [tex]d = v * t[/tex]

Now from the question we see that d is the same for both sound waves

 So

        [tex]v_c t = v_s * t_1[/tex]

Now  

So [tex]t_1 = k + t[/tex]

      [tex]v_c t = v_s * (t+ k)[/tex]

=>     [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]t = 0.0645 \ s[/tex]

So

     [tex]d = 3000 * 0.0645[/tex]

     [tex]d = 193.6 \ m[/tex]

       

     

A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film

Answers

Answer:

t(min) = 94nm

Explanation:

The wavelength of the light incident is 500 nm.

The refractive index of the film is 1.33.

The minimum thickness of the soap film required for constructive interference.

The thickness of the film for the constructive interference is given by:

2*t= (m + 1/2) λ′

Now, λ′ = λ/μ = 500/1.33 = 376nm

The minimum thickness of the film

′t′ will be at m=0 :

2*t(min) = (0 + 1/2) 376

t(min) = 94nm

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.

Required:
What is the charge on the dust particle?

Answers

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

   

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

The following equation is an example of
decay.
181
185
79
Au →
4
2
He+

Answers

Answer:

Alp decay.

Explanation:

From the above equation, the parent nucleus 185 79Au produces a daughter nuclei 181 77 Ir.

A careful observation of the atomic mass of the parent nucleus (185) and the atomic mass of the daughter nuclei (181) shows that the atomic mass of the daughter nuclei decreased by a factor of 4. Also, the atomic number of the daughter nuclei also decreased by a factor of 2 when compared with the parent nucleus as shown in the equation given above.

This simply means that the parent nucleus has undergone alpha decay which is represented with a helium atom as 4 2He.

Therefore, the equation is an example of alpha decay.

If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units?

Answers

Answer:

The mas unit is the the 'Kilogram' written as 'kg'

Volume is 10 L

Explanation:

The complete question is

If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?

Second, what is the volume

mass units is the 'Kilogram', written as 'kg'

density = mass/volume = 100 kg/L

the mass  = 1000 kg

volume = mass/density = 1000/100 = 10 L

A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm from the screen, the first-order maxima appear 6.09 cmcm from the center of the pattern. How many lines per millimeter does this grating have?

Answers

Answer:

221 lines per millimetre

Explanation:

We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.

Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,

tanθ = 6.09 cm/42.0 cm = 0.145

From trig ratios, cot²θ + 1 = cosec²θ

cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969

sinθ = 1/cosecθ = 1/6.969 = 0.1435

Also, sinθ = mλ/d at the first-order maximum, m = 1. So

sinθ = (1)λ/d = λ/d

Equating both expressions we have  

0.1435 = λ/d

d = λ/0.1435

Now, λ = 650 nm = 650 × 10⁻⁹ m

d = 650 × 10⁻⁹ m/0.1435

d = 4529.62 × 10⁻⁹ m per line

d = 4.52962 × 10⁻⁶ m per line

d = 0.00452962 × 10⁻³ m per line

d = 0.00452962 mm per line

Since d = width of grating/number of lines of grating

Then number of lines per millimetre = 1/grating spacing

= 1/0.00452962

= 220.77 lines per millimetre

≅ 221 lines per millimetre since we can only have a whole number of lines.

A uniform disk with mass 43.9 kgkg and radius 0.280 mm is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.0 NN is applied tangent to the rim of the disk. Part A What is the magnitude vv of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.400 revolution

Answers

Answer:

1.36 m/s

Explanation:

I = ½mR²

τ = FR

α = τ/I = FR / (½mR²) = 2F/mR

a = Rα = 2F/m

s = θR

v² = u² + 2as

u = 0

v = √2as = √(2(2F/m)(θR)) = 2√(FθR/m)

v = 2√(29.0(0.400)(2π)(0.280) / 43.9) = 1.3636272...

Red and orange stars are found evenly spread throughout the galactic disk, but blue stars are typically found

Answers

Answer:

only in or near star-forming clouds

Explanation:

When in the galactic disk, Red and orange stars are found evenly spread so here Blue stars are hot and therefore massive and therefore short-lived,  that is means they never have time to venture far from the places, where they were born. so correct answer is blue stars are typically found only in or near star-forming clouds

How much energy is required to accelerate a spaceship with a rest mass of 121 metric tons to a speed of 0.509 c?

Answers

Answer

1.07E22 Joules

Explanation;

We know that mass expands by a factor

=>>1/√[1-(v/c)²]

But v= 0.509c

So

1/√(1 - 0.509²)

=>>> 1/√(1 - 0.2591)

= >> 1/√(0.7409) = 1.16

But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy

And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.

So Multiplying by 119.84

Kinetic energy will be 1.07x 10^22 joules

A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?

Answers

Answer:

There is no induced current on the coil.

Explanation:

Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.

Light of wavelength 520 nm is incident a on a diffraction grating with a slit spacing of 2.20 μm , what is the angle from the axis for the third order maximum?

Answers

Answer:

θ = 45.15°

Explanation:

We need to use the grating equation in this question. The grating equation is given as follows:

mλ = d Sin θ

where,

m = order number = 3

λ = wavelength of light = 520 nm = 5.2 x 10⁻⁷ m

d = slit spacing = 2.2 μm = 2.2 x 10⁻⁶ m

θ = angle from the axis = ?

Therefore,

(3)(5.2 x 10⁻⁷ m) = (2.2 x 10⁻⁶ m) Sin θ

Sin θ = (3)(5.2 x 10⁻⁷ m)/(2.2 x 10⁻⁶ m)

Sin θ = 0.709

θ = Sin⁻¹(0.709)

θ = 45.15°

A long solenoid of radius 3 cm has 1100 turns per meter. If the solenoid carries a current of 1.5 A, then calculate the magnetic field at the center of the solenoid.a. 2.1E^-3T b. 1.0E^-3 T c. 1.7E^-4T d. 7.0E^-2 T

Answers

Answer:

The magnetic field at the center of the solenoid is 2.1  × 10⁻³ T

Explanation:

The magnetic field B at the center of the solenoid is given by

B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.

So B = μ₀ni

= 4π × 10⁻⁷H/m × 1100 × 1.5 A

= 4π × 10⁻⁷H/m × 1650 A-turns/m

= 20734.5 × 10⁻⁷T  

= 2.07345 × 10⁻³ T

≅ 2.1  × 10⁻³ T

So the magnetic field at the center of the solenoid is 2.1  × 10⁻³ T

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.

Answers

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]

Where

[tex]\lambda[/tex] represents wavelength

R represents Rydberg's constant

[tex]n_f[/tex] represents Final energy states

and [tex]n_i[/tex] represents initial energy states

Now Substitute is

[tex]1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]

now we will put the values into the above formula

[tex]= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )[/tex]

[tex]= 1218888.889 m^{-1}[/tex]

Now we will rewrite the answer in the term of [tex]\lambda[/tex]

[tex]\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m[/tex]

So, the whole Paschen series is in the part of the spectrum.

What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?​

Answers

Answer:

If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.

The simple reason is the desirability for genetic variation using recessive genes.

In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.

Hope this helps

A cyclist moves effortlessly at a constant speed of 12 m / s, but enters a muddy area where the coefficient of kinetic friction is 0.60. Will the rider leave the muddy area without having to pedal if the mud extends 11m? If so, how fast will it emerge?

Answers

Answer:

3.5 m/s

Explanation:

There are 3 forces on the cyclist:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

∑F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

a = -(10 m/s²)(0.60)

a = -6 m/s²

Velocity reached at end of 11 m:

v² = v₀² + 2aΔx

v² = (12 m/s)² + 2 (-6 m/s²) (11 m)

v = √12 m/s

v ≈ 3.5 m/s

coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second later. assume g is 10m/s. how far below the top of the tower des coin 1 pass coin 2

Answers

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

the height below the tower at which coin 1 passes coin 2

Find the maximum height attained by coin 1 before falling to the ground:

[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]

But the time taken for the coin 1 to reach 11.25 m above the tower:

[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]

The time  at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

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A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm

Answers

Explanation:

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The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

What is diffraction?

Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.

Given:

The number of lines = 6000 per cm,

The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m

Calculate the diffraction grating,

[tex]d = 1 / no\ of\ lines[/tex]

d = 10⁻² / 6000 m,

Calculate the second-order maxima angle and third-order maxima angle by the formula given below,

[tex]dsin\theta_1 = n_1 \lambda[/tex]

[tex]sin\theta_1 = n_1\lambda / d[/tex]

[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]

θ₁ = sin⁻¹(0.6)

θ₁ = 36.87°

Similarly, for θ₂,

θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)

θ₂ = sin⁻¹(0.9)

θ₂ = 64.16°

Calculate the separation as follows,

θ₂ - θ₁ = 64.16° - 36.87°

θ₂ - θ₁ =  27.29°

Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

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A mass M slides downward along a rough plane surface inclined at angle \Theta\:Θ= 32.51 in degrees relative to the horizontal. Initially the mass has a speed V_0\:V 0 = 6.03 m/s, before it slides a distance L = 1.0 m down the incline. During this sliding, the magnitude of the power associated with the work done by friction is equal to the magnitude of the power associated with the work done by the gravitational force. What is the coefficient of kinetic friction between the mass and the incline?

Answers

Answer: μ = 0.8885

Explanation: Force due to friction is calculated as: [tex]F_{f} = \mu.N[/tex]

At an inclined plane, normal force (N) is: N = mgcosθ, in which θ=32.51.

Power associated with work done by friction is [tex]P=F_{f}.x[/tex]. The variable x is displacement the object "spent its energy".

Power associated with work done by gravitational force is P = mghcosθ, where h is height.

The decline forms with horizontal plane a triangle as draw in the picture.

To determine force due to friction:

[tex]F_{f}.x=mghcos(\theta)[/tex]

[tex]F_{f}=\frac{mghcos(\theta)}{x}[/tex]

Replacing force:

[tex]\frac{m.g.h.cos(\theta)}{x} = \mu.m.g.cos(\theta)[/tex]

[tex]\mu=\frac{h}{x}[/tex]

Calculating h using trigonometric relations:

[tex]sin(32.51) = \frac{h}{1}[/tex]

h = sin(32.51)

Coefficient of Kinetic friction is

[tex]\mu=\frac{sin(32.51)}{1}[/tex]

μ = 0.8885

For these conditions, coefficient of kinetic friction is μ = 0.8885.

A laser beam takes 15 ms to travel from a rocket to the reflective surface of a planet and back to the rocket. How far is the rocket from this planet's surface

Answers

Explanation:

It took 7.5 ms or [tex]7.5×10^{-3}\:\text{s}[/tex] for the laser beam to reach the planet's surface. Since the speed of light is [tex]3×10^8\:\text{m/s}[/tex], the distance is

[tex]d = vt = (3×10^8\:\text{m/s})(7.5×10^{-3}\:\text{s})[/tex]

[tex]\:\:\:\:=2.25×10^{6}\:\text{m}[/tex]

A wooden ice box has a total area of 1.50 m2 amd walls with an average thickness of 2.0 cm. The box contains ice at 0.0 oC. The inside of the box is kept cold by melting ice. How much ice melts in one day if the ice box is kept in the shade of tree at 29 oC. (Assume the thermal conductivity of wood is 0.16 kJ/s m oC

Answers

Answer:

m = 9.1 x 10⁶ kg

Explanation:

First, we need to find the rate of heat transfer through the box to the ice. For this purpose, we use Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

Q = Rate Of Heat Transfer = ?

K = Thermal Conductivity = 0.16 KW/m.°C = 160 W/m.°C

A = Area = 1.5 m²

ΔT = Difference in Temperature = 29°C - 0°C = 29°C

L = Thickness of wall = 2 cm = 0.002 m

Therefore,

Q = (160 W/m °C)(1.5 m²)(29°C)/(0.002 m)

Q = 3.48 x 10⁶ W

Now, we find the amount of heat transferred in one day to the ice:

q = Qt

where,

q = amount of heat = ?

t = time = (1 day)(24 h/1 day)(3600 s/1 h) = 86400 s

Therefore,

q = (3.48 x 10⁶ W)(8.64 x 10⁴ s)

q = 3 x 10¹¹ J

Now, for mass of ice melted in a day:

q = m H

m = q/H

where,

m = mass of ice melted in a day = ?

H = latent heat of fusion of ice = 3.3 x 10⁵ J/kg

Therefore,

m = (3 x 10¹¹ J)/(3.3 x 10⁵ J/kg)

m = 9.1 x 10⁶ kg

?a wire is stretched 30% what is the percentage change in resistance ​

Answers

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]

So that;

[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])

    = (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])

    = [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]

   = 1.69R               (∵ R = (ρl) ÷ A)

[tex]R_{o}[/tex] = 1.69R

Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.

The change in resistance of the wire = [tex]R_{o}[/tex] - R

                                      = 1.69R  - 1R

                                      = 0.69R

The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100

                                                               = 0.69 × 100

                                                              = 69%

The percentage change in resistance of the wire is 69%.

A/An is a type of blood cell that's also called a red blood cell. a) Jeukocyte O b) thrombocyte c) plasma d) erythrocyte

Answers

Answer:

red blood cell, also called erythrocyte

Explanation:

Hope it helps

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A 0.100-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.550 m apart and 2.0 m long,
(a) If the coefficient of kinetic friction between the rod and rails is 0.120, what vertical magnetic field is required to keep the rod moving at a constant speed?
(b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).

Answers

Answer:

The speed of the rod is 2.169 m/s.

Explanation:

Given that,

Mass = 0.100 kg

Current = 15.0 A

Distance = 2 m

Length = 0.550 m

Kinetic friction = 0.120

(a). We need to calculate the magnetic field

Using relation of frictional force and magnetic force

[tex]F_{f}=F_{B}[/tex]

[tex]\mu mg=Bli[/tex]

[tex]B=\dfrac{\mu mg}{li}[/tex]

Where, l = length

i = current

m = mass

Put the value into the formula

[tex]B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}[/tex]

[tex]B=0.01425\ T[/tex]

[tex]B=1.425\times10^{-2}\ T[/tex]

(b). If the friction between the rod and rail is reduced zero.

So, [tex]f_{f}=0[/tex]

We need to calculate the acceleration

Using formula of force

[tex]F_{net}=f_{f}+F_{B}[/tex]

[tex]F_{net}=0+Bil[/tex]

[tex]ma=Bil[/tex]

[tex]a=\dfrac{Bil}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}[/tex]

[tex]a=1.176\ m/s^2[/tex]

We need to calculate the speed of the rod

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Put the value into the formula

[tex]v^2=0+2\times1.176\times2[/tex]

[tex]v^2=\sqrt{4.704}\ m/s[/tex]

[tex]v=2.169\ m/s[/tex]

Hence, The speed of the rod is 2.169 m/s.

21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fricción μr =0.55. ¿Cuánto desciende antes de detenerse?

Answers

Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

             

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

         

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

Determine the point of contraflexure

Answers

Answer:

The point of contraflexure (PoC) occurs where bending is zero and at the point of change between positive and negative (or between compression and tension). In a beam that is flexing (or bending), the point where there is zero bending moment is called the point of contraflexure.

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