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 1

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°


Related Questions

A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The distance it moves in the third second.

Answers

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= - 4 m/s

A wire of 5.8m long, 2mm diameter carries 750ma current when 22mv potential difference is applied at its ends. if drift speed of electrons is found then:_________.
(a) The resistance R of the wire(b) The resistivity p, and(c) The number n of free electrons per unit volume.​

Answers

Explanation:

According to Ohms Law :

V = I * R

(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms

Also,

[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]

(B)

[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]

Drift speed is missing. It is given as;

1.7 × 10^(-5) m/s

A) R = 0.0293 ohms

B) ρ = 1.589 × 10^(-8)

C) n = 8.8 × 10^(28) electrons

This is about finding, resistance and resistivity.

We are given;

Length; L = 5.8 m

Diameter; d = 2mm = 0.002 m

Radius; r = d/2 = 0.001 m

Voltage; V = 22 mv = 0.022 V

Current; I = 750 mA = 0.75 A

Area; A = πr² = 0.001²π

Drift speed; v_d = 1.7 × 10^(-5) m/s

A) Formula for resistance is;

R = V/I

R = 0.022/0.75

R = 0.0293 ohms

B) formula for resistivity is given by;

ρ = RA/L

ρ = (0.0293 × 0.001²π)/5.8

ρ = 1.589 × 10^(-8)

C) Formula for current density is given by;

J = n•e•v_d

Where;

J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²

e is charge on an electron = 1.6 × 10^(-19) C

v_d = 1.7 × 10^(-5) m/s

n is number of free electrons per unit volume

Thus;

238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))

238732.44 = (2.72 × 10^(-24))n

n = 238732.44/(2.72 × 10^(-24))

n = 8.8 × 10^(28)

Read more at; brainly.com/question/17005119

When light of wavelength 233 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.98 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface

Answers

Answer:

λmax = 372 nm

Explanation:

First we find the energy of photon:

E = hc/λ

where,

E = Energy of Photon = ?

λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m

c = speed of light = 3 x 10⁸ m/s

h = Planks Constant = 6.626 x 10⁻³⁴ J.s

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)

E = 8.5 x 10⁻¹⁹ J

Now, from Einstein's Photoelectric Equation:

E = Work Function + Kinetic Energy

8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)

Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J

Work Function = 5.332 x 10⁻¹⁹ J

Since, work function is the minimum amount of energy required to emit electron. Therefore:

Work Function = hc/λmax

λmax = hc/Work Function

where,

λmax = maximum wavelength of light that will produce photoelectrons = ?

Therefore,

λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)

λmax = 3.72 x 10⁻⁷ m

λmax = 372 nm

Calculate the focal length (in m) of the mirror formed by the shiny bottom of a spoon that has a 3.40 cm radius of curvature. m (b) What is its power in diopters? D

Answers

Answer:

The power of the mirror in diopters is 58.8 D

Explanation:

Given;

radius of curvature of the spoon, R = 3.4 cm = 0.034 m

The focal length of a mirror is given by;

[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]

The focal length of the mirror is 0.017 m

(b) The power of the mirror is given by;

[tex]P = \frac{1}{f}[/tex]

where;

P is the power of the mirror

f is the focal length

[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]

Thus, the power of the mirror in diopters is 58.8 D

An object is inside a room that has a constant temperature of 289 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object

Answers

Answer:

T_object = 380.35 K

Explanation:

From Stefan–Boltzmann law, the power output is given by the formula:

P = σAT⁴

where;

σ is Stefan-Boltzmann constant

A is area of the radiating surface.

T is temperature of the body

Now, we are told that the power the object emitted is 3 times the power absorbed from the room.

Thus, we have;

P_e = 3P_a

Where P_e is power emitted and P_a is power absorbed.

So, we have;

σA(T_object)⁴ = 3σA (T_room)⁴

σA will cancel out to give;

(T_object)⁴ = 3(T_room)⁴

We are given T_room = 289 K

Thus;

(T_object)⁴ = 3 × 289⁴

(T_object) = ∜(3 × 289⁴)

T_object = 380.35 K

Please help!
Much appreciated!​

Answers

Answer:

F = 2.7×10¯⁶ N.

Explanation:

From the question given:

F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²

Thus we can obtain the value value of F by carrying the operation as follow:

F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²

F = 2.7648×10¯⁷ / 0.1024

F = 2.7×10¯⁶ N.

Therefore, the value of F is 2.7×10¯⁶ N.

Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?

The direction for each field vector is perpendicular to equipotential lines.

Take a snapshot of the simulation showing equipotential lines and paste to a word document.

Answers

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An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s

Answers

Answer:

The charge is  [tex]Q = 2.177 *10^{-9} \ C[/tex]

Explanation:

From the question we are told that

     The electric potential is  [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]

     The  mass of the drop is  [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]

      The  speed is  [tex]v = 18.8 \ m/s[/tex]

Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question  is mathematically represented as

       [tex]Q = \frac{m v^2 }{ 2 * V }[/tex]

Substituting values

      [tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]

       [tex]Q = 2.177 *10^{-9} \ C[/tex]

what effect does decreasing the field current below its nominal value have on the speed versus voltage characteristic of a separately excited dc motor

Answers

Answer

The effect is that it Decreases the field current IF and increases slope K1

An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave travels at 5 000 m/s. If P-waves are received at a seismic station 1.00 minute before an S-wave arrives, how far away is the earthquake center?

Answers

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s

Answers

(a) The car is undergoing an acceleration of

[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]

so that in 9.02 s, it will have covered a distance of

[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]

The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:

[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]

(b) The wheels have average angular velocity

[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]

where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.

The wheels start at rest, so

[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?

Answers

Answer:

The value is  [tex]\mu = 0.76[/tex]

Explanation:

From the question we are told that

    The  acceleration is [tex]a = 7.4 \ m /s^2[/tex]

Generally the force by which the truck bed (truck) is moving with is mathematically represented as

          [tex]F = ma[/tex]

Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet  is equal the force by which the the truck bed is moving with that is  

        [tex]F_f = F[/tex]

Here  [tex]F_f[/tex] is the frictional force which is mathematically represented as

         [tex]F_f = \mu * m * g[/tex]

substituting into above equation

         [tex]\mu * m * g = ma[/tex]

=>        [tex]\mu = \frac{a}{g}[/tex]

substituting values

           [tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]

           [tex]\mu = 0.76[/tex]

         

If a soap bubble is 130 nmnm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated normally by white light

Answers

Answer:

The question is not complete, here is the other part.

Assume that n = 1.36.

Express your answer to three significant figures and include the appropriate units.

λ = 707.2nm

Explanation:

n = 1.36

t = 130

2 n t= (m+1/2) λ

To solve for λ.

λ = 2 n t ÷ m+1/2

λ = 2 × 1.36 × 130 ÷ m +1/2

λ =

If m= 0

λ= 353.6 ÷ 0+ 1/2

λ = 353.6 × 2

= 707.2nm

A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?Please help i will give brainliest!!!! lower the board so it's level with the ground ____ roughen up the texture of the wooden board ___ raise the board to a higher angle ____ press down on the brick in a direction that is perpendicular to the board _____

Answers

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

       fr - Wₓ = m    a                      (1)

Y axis  

       N- [tex]W_{y}[/tex] = 0

       N = W_{y}

let's use trigonometry to find the components of the weight

        sin θ = Wₓ / W

        cos θ = W_{y} / W

        Wₓ = W sin θ

        W_{y} = W cos θ

the friction force has the formula

         fr = μ N

         fr = μ Wy

         fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

           F = fr

           F = (μ  mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

To a person swimming 0.80 m below the water surface of a swimming pool, the diving board directly overhead appears to be a height of 5.20 m above the swimmer. What is the actual height of the diving board above the water surface

Answers

Answer:

The actual height is 3.308 m.

Explanation:

The person is swimming below the water surface at distance = 0.80 m  

The height of the diving board appears at a distance or height = 5.20 m

Now we have to find the actual distance of the diving board from the water surface.

We know the refractive index of water is 1.33.

Therefore, the actual height = (Distance that appears – distance below the water surface) / Refractive index.

The actual height = ( 5.20 - 0.80 ) / 1.33 = 3.308 m

How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?

Answers

Answer:

a) the distance between the interference fringes is reduced by half

b) the distance between stripes is doubled

Explanation:

Interference experiments constructive interference is described by the expression

          d sin θ = m λ

let's use trigonometry to find the distance between the interference fringes

              tan θ=  y / L

dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles

            tan θ = sin θ / cos θ = sin θ

we substitute

             sin θ = y / L

             

            d y / L = m  λ

           y = m λ / d L

a) it asks us when the screen doubles its distance

           L ’= 2 L

subtitute in the equation

           y ’= m λ / (d 2L)

           y ’=( m λ / d L) /2

           y ’= y / 2

the distance between the interference fringes is reduced by half

b) the density of the network doubles

      if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half

            d ’= d / 2

we substitute

          y ’= m λ (L d / 2)

          y ’= m λ / (L d) 2

          y ’= y 2

the distance between stripes is doubled

The entropy of any substance at any temperature above absolute zero is called the: Select the correct answer below:
a. absolute entropy
b. Third Law entropy
c. standard entropy
d. free entropy
e. none of the above

Answers

Answer:

b. Third Law entropy  

Explanation:

Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature"  tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.

In the question above, the correct answer is option b.

A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance

Answers

Answer:

8 mm

Explanation:

From the information given:

The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.

when the distance is less than the radius ; we have:

[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]

when the distance is greater than the radius; we have:

[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]

Equating both equations together ; we have :

[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]

[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]

[tex]R= \dfrac{d^2}{r}[/tex]

where; d = radius of the wire and r = distance;

[tex]R =\dfrac{4^2}{2}[/tex]

[tex]R =\dfrac{16}{2}[/tex]

R = 8 mm

A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface

Answers

Answer:

The radius is  [tex]r = 3.1905 \ m[/tex]

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  [tex]d = 2.70 \ m[/tex]

        The refractive index of the liquid is  [tex]n_i = 1.31[/tex]

Now the critical value is mathematically represented as

         [tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]

substituting values

         [tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]

         [tex]\theta = 49.76^o[/tex]

Using SOHCAHTOA rule we have that

         [tex]tan \theta = \frac{ r}{d}[/tex]

=>     [tex]r = d * tan \theta[/tex]

substituting values  

        [tex]r = 2.7 * tan (49.76)[/tex]

        [tex]r = 3.1905 \ m[/tex]

         

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?

Answers

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

We have that for the Question, it can be said that the average induced emf in the coil is

E=0.028565V

From the question we are told

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?

Generally the equation for the Average emf induced   is mathematically given as

[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]

Hence

[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]

Therefore

[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]

E=0.028565V

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Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?

Answers

Answer:

a

   [tex]\lambda = 1.667 nm[/tex]

b

     [tex]\theta = 0.8681^o[/tex]

Explanation:

From the question we are told that

   The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]

    The  is distance of the screen from the slit is  [tex]D = 2.60 \ m[/tex]

    The distance between the central bright fringe and either of the adjacent bright   [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]

Generally  the condition for constructive interference is  

      [tex]d sin \tha(\theta ) = n \lambda[/tex]

From the question we are told that small-angle approximation is valid here.

So    [tex]sin (\theta ) = \theta[/tex]

=>        [tex]d \theta = n \lambda[/tex]

=>        [tex]\theta = \frac{n * \lambda }{d }[/tex]

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         [tex]y = D * sin (\theta )[/tex]

From the question we are told that small-angle approximation is valid here.

So

       [tex]y = D * \theta[/tex]

=>   [tex]\theta = \frac{ y}{D}[/tex]

So

     [tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]

     [tex]\lambda =\frac{d * y }{n * D}[/tex]

substituting values

       [tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]

        [tex]\lambda = 1.667 *10^{-6}[/tex]

        [tex]\lambda = 1.667 nm[/tex]

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

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

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

    [tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]

    [tex]\theta = 0.8681^o[/tex]

Rank the ultraviolet, infrared, and visible regions of theelectromagnetic spectrum in terms of lowest to highest energy,frequency, and wavelength.
Energy: < <
Frequency: < <
Wavelength: <

Answers

Answer:

1. Energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared >> visible > ultraviolet

Explanation:

Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.

Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:

1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.

i.e energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.

i.e frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.

i.e  wavelength: infrared >> visible > ultraviolet

In terms of lowest to the highest energy,frequency, and wavelength is;

Energy: infrared > visible light > ultraviolet

Frequency: infrared > visible light > ultraviolet

Wavelength: ultraviolet > visible light > infrared

The electromagnetic spectrum:

The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.

The ultraviolet: This wave is seen in the sunlight and is made up of wavelength of 10nm to 400nm. A frequency of [tex]10^{16}[/tex](Hz).Infrared wave: They are invisisble to the human eye but can be felt as heat. It has frequency of [tex]10^{12}[/tex]Hz and a wavelength of 780nm to 1mm.Visible light: This is part of the electromagnetic wave that the eye can view. It has frequency of [tex]10^{15}[/tex]Hz and a wavelength of 380 to 700nm.

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In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.497 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen

Answers

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m

For an object to move, a(n) _______ force must be applied. Question 1 options: Balanced Unbalanced

Answers

Answer:

Unbalenced

Explanation:

when balenced forces are applied to an object there is no motion. When you apply unbalenced force the object you are applying the force to will move in the opposite direction of the force.

Answer:

im pretty sure it unbalenced

Explanation:

i just am

A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish

Answers

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

The apparent depth of the fish is 60 cm.

To calculate the apparent depth of the fish, we use the formula below.

Formula:

R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1

Where:

R.F = Refractive index of water

Make D' The subject of the equation.

D' = D/R.F................... Equation 2

From the question,

Given:

D = 80 cmR.F = 1.333

Substitute these values into equation 2

D' = 80/1.33D' = 60.01D' = 60 cm

Hence, the apparent depth of the fish is 60 cm

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The positron has the same mass as an electron, with an electric charge of +e. A positron follows a uniform circular motion of radius 5.03 mm due to the force of a uniform magnetic field of 0.85 T. How many complete revolutions does the positron perform If it spends 2.30 s inside the field? (electron mass = 9.11 x 10-31 kg, electron charge = -1.6 x 10-19 C)

Answers

Answer:

5.465 × 10^10 revolutions

Explanation:

Formula for Magnetic Field = m. v/ q . r

M = mass of electron = mass of positron = 9.11 x 10^-31 kg,

radius of the positron = 5.03 mm

We convert to meters.

1000mm = 1m

5.03mm = xm

Cross multiply

x = 5.03/1000mm

x = 0.00503m

q = Electric charge = -1.6 x 10^-19 C

Magnetic field (B) = 0.85 T

Speed of the positron is unknown

0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503

0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v

v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg

v = 6.8408 ×10-22/ 9.11 x 10^-31 kg

v = 750911086.72m/s

Formula for complete revolutions =

Speed × time / Circumference

Time = 2.30s

Circumference of the circular path = 2πr

r =0.00503

Circumference = 2 × π × 0.00503

= 0.0316044221

Revolution = 750911086.72 × 2.30/0.0316044221

= 1727095499.5/0.0316044221

= 546541562294 revolutions

Approximately = 5.465 × 10^10 revolutions

A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary

Answers

Answer:

12A

Explanation:

Formula for calculating the relationship between  the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:

[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]  where;

Vs and Vp are the emf in the secondary and primary coil respectively

Ns and Np are the number if turns in the secondary and primary coil respectively

Ip and Is are the currents in the secondary and primary coil respectively

Since the are all equal to each other, then we can equate any teo of the expression as shown;

[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

Given parameters

Np = 500-turns

Ns = 2000-turns

Is = 3.0Amp

Required

Current in the primary coil (Ip)

Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]

[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]

Hence the current in the primary coil is 12Amp

Your favorite radio station broadcasts at a frequency of 91.5 MHz with a power of 11.5 kW. How many photons does the antenna of the station emit in each second?

Answers

Answer:

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Explanation:

Given:

Frequency = 91.5 MHz

Power = 11.5 Kw = 11,500 J/s

Find:

Number of photons emit per second

Computation:

Total energy with frequency (E) = hf

Total energy with frequency (E) = 6.626×10⁻³⁴  × 91.5×10⁶

Total energy with frequency (E) = 6.06×10⁻²⁶ J

Number of photons emit per second = 11,500 / 6.06×10⁻²⁶

Number of photons emit per second = 1897.689 × 10²⁶

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Based on The MOHS hardness Scale, which mineral could be scratched by a penny but not by a fingernail

A. Fluorite
B. Calcite
C. Gypsum
D. Talc

Answers

The correct answer is B. Calcite

Explanation:

Mohs hardness scale indicates the hardness of minerals using a scale from 1 to 10 as well as defining the objects or tools that can be used to scratch the minerals. These two features of minerals are shown in the table of the image. About this, it is shown gypsum and talc can be scratched by just a fingernail, considering minerals with a hardness of 2.5 or below can be scratched by a fingernail. In the case of calcite that has a hardness of 3, this cannot be scratched by a fingernail, but it can be scratched by a penny, which works for minerals with a hardness of 3.5 or below. Thus, the correct answer is Calcite.

What is the force that attracts objects with mass toward each other?

Answers

Explanation:

gravitional force attracts objects with mass toward each other.

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