A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light

Answers

Answer 1

Answer:

b

Explanation:


Related Questions

"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit

Answers

Answer:

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Explanation:

Given that:

the time period T = 88.5 min = 88.5 × 60  = 5310 sec

The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg

if  the radius of orbit is r,

Then,

[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e }{r}[/tex]

[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Similarly :

[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]

where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Then:

[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]

[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]

[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]

[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]

[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]

[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]

[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]

[tex]r= 2565.38^2[/tex]

r = 6579225 m

The  circumference of the satellites  orbit can now be determined by using the formula:

 circumference = 2π r

 circumference = 2π  × 6579225 m

 circumference = 41338489.85 m

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.

Answers

Answer:

they must be affordable because they have to pay for it or they wont get the stuff they are bying.

Explanation:

need a brainliest please.

Answer: B, they must be affordable.

Explanation:

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.


Required:

a. What is its frequency?

b. What type of electromagnetic radiation might this be?

Answers

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin. Part A What is the maximum wavelength of electromagnetic radiation that can cause a transition

Answers

Answer:

The maximum wavelength of the e-m wave is 6.9 x 10^-7 m

Explanation:

Energy required to trigger a response = 1.8 eV

we convert to energy in Joules.

1 eV = 1.602 x 10^-19 J

1.8 eV = [tex]x[/tex] J

[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J

The energy of an electromagnetic wave is gotten as

E = hf

where

h is the Planck's constant = 6.63 x 10^-34 J-s

and f is the frequency of the wave.

substituting values, we have

2.88 x 10^-19 = 6.63 x 10^-34 x f

f = (2.88 x 10^-19)/(6.63 x 10^-34)

f = 4.34 x 10^14 Hz

We know that the frequency of an e-m wave is given as

f = c/λ

where

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength of the e-m wave

From this we can say that

λ = c/f

λ = (3 x 10^8)/(4.34 x 10^14)

λ = 6.9 x 10^-7 m


I MIND TRICK PLZ HELP LOL
Troy and Abed are running in a race. Troy finishes the race in 12 minutes. Abed finishes the race in 7 minutes and 30 seconds. If Troy is running at an average speed of 3 miles per hour and speed varies inversely with time, what is Abed’s average speed for the race?

Answers

Answer:

Explanation:

Let the race be of a fixed distance x

[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]

Troy's Average speed = 3 miles/hr = x / 0.2 hr

x = 0.6 miles

Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion

Answers

Answer:

C. planetary accretion

Explanation:

Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.

Answer:

[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]

Explanation:

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.

Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.

A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black c. White d. Green

Answers

Red light reflects off the card into your eyes and you see the red card as red. The light will just make the card brighter. So A

The color that is reflected when a red card is illuminated by red light is white.

The color an object is perceived to have, depends on the frequency of light it reflects.

If white light incidents on a red filter, red is transmitted while blue and green are absorbed.

Consequently, when a red card is illuminated by red light, the red card will  reflect back almost all the incident light on it, causing it to appear brighter which creates an  illusion of white color to the eyes.

Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.

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The mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal. (1) Write an equation that represents this direct proportion, in which D is the proportionality constant. The density of lead metal is 11.3 g/cm3. (2) What is the mass of a piece of lead metal that has a volume of 17.3 cm3

Answers

Answer:

1) M = 11.3V2) 195.49 grams

Explanation:

1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;

M ∝ V

M = kV

For every proportionality sign, there will always be a proportionality constant 'k'

Since the proportionality constant is the density (D) of the metal, the equation will become;

M = DV

Given the density to be 11.3 g/cm3, the equation will become;

M = 11.3V

Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V

2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;

M = 11.3V

M = 11.3 * 17.3

M = 195.49 grams

Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.

A ball is thrown upward from a height of 432 feet above the​ ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. ​a) Find​ s(t), the function giving the height of the ball at time t. ​b) How long will the ball take to reach the​ ground? ​c) How high will the ball​ go?

Answers

Answer;

A)S(t)=96t-16t² +432

B)it will take 9 seconds for the ball to reach the ground.

C)864feet

Explanation:

We were given an initial height of 432 feet.

And v(t)= 96-32t

A) we are to Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

S(t)= ∫(96-32)dt

S(t)=96t-16t² +K

S(t)=96t-16t² +432

In which the constant of integration K is the initial height, so K= 432

b) we need to know how long will the ball take to reach the​ ground

This is t when S(t)= 0

S(t)=96t-16t² +432

-16t² +96t +432=0

This is quadratic equation, if you solve using factorization method we have

t= -3 or t= 9

Therefore, , t is the instant of time and it must be a positive value.

So it will take 9 seconds for the ball to reach the ground.

C)V=s/t

Velocity= distance/ time

=96=s/9sec

S=96×9

=864feet

By applying the integrations,

(a) [tex]S = 96t-16t^2+432[/tex]

(b) Time will be "t = 9".

(c) Height will be "576"

Given:

Height,

423 feet

Initial velocity,

96 feet/sec

According to the question,

(a)

Integrate v:

[tex]S = 96t-16t^2+C[/tex]

Initial Condition,

→ [tex]S = 96t-16t^2+432[/tex]

(b)

Hits the ground when,

S = 0

→ [tex]0=96t-16t^2+432[/tex]

→ [tex]t =9[/tex]

(c)

Maximum height when,

v = 0

→ [tex]0 = 96-32 t[/tex]

→ [tex]t = 3[/tex]

Now,

→ [tex]S = 96\times 3-16\times 3^2+432[/tex]

      [tex]= 576[/tex]

Thus the answer above is correct.

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Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?

Answers

Answer:

Kindly check explanation

Explanation:

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.

Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

The calculation is as follows;

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

[tex](1093.613 \times x) = 45[/tex]

[tex]x = 45 \div 1093.613[/tex]

x = 0.0411480 km

here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.

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(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]

For more information on potential difference visit

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Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.

Answers

The researchers need to compare those who contracted the disease to those who did not.

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]

NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?

Answers

Answer:

A = 6.8 km²

Explanation:

A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.

B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;

F_rad = 2IA/c

I is given by the formula;

I = P/(4πr²)

Thus;

F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²

Where;

A is the area of the sail

r is the distance of the sail from the sun

c is the speed of light = 3 × 10^(8) m/s

P is total power output of the sun = 3.90 × 10^(26) W

Now,F_rad = F_g

Where F_g is gravitational force.

Thus;

PA/2cπr² = G•m•M_sun/r²

r² will cancel out to givw;

PA/2cπ = G•m•M_sun

Making A the subject, we have;

A = (2•c•π•G•m•M_sun)/P

Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg

G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²

Thus;

A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))

A = 6.8 × 10^(6) m² = 6.8 km²

A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

Answers

Answer:

The moment of inertia is  [tex]I= 312.09 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

    The  mass of the platform is   m =  137 kg

     The radius is  r  =  1.53 m

    The mass of the person is  [tex]m_p = 68.7 \ kg[/tex]

    The distance of the person from the center is  [tex]d_c =1.19 \ m[/tex]

    The mass of the dog is  [tex]m_d = 25.9 \ kg[/tex]

     The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]

Generally the moment of inertia of the system is mathematically represented as

      [tex]I = I_1 + I_2 + I_3[/tex]

Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as

          [tex]I_1 = \frac{m * r^2}{2}[/tex]

substituting values

           [tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]

           [tex]I_1 = 160.35 \ kg\cdot m^2[/tex]

Also  [tex]I_2[/tex]  is the moment of inertia of the person about the axis which is mathematically represented as

          [tex]I_2 = m_p * d_c^2[/tex]

substituting values

          [tex]I_2 = 68.7 * 1.19^2[/tex]

          [tex]I_2 = 97.29 \ kg \cdot m^2[/tex]

Also  [tex]I_3[/tex]  is the moment of inertia of the dog about the axis which is mathematically represented as

          [tex]I_3 = m_d * d_d^2[/tex]

substituting values

          [tex]I_3 = 25.9 * 1.45^2[/tex]

          [tex]I_3 = 54.45 \ kg \cdot m^2[/tex]

Thus  

        [tex]I= 160.35 + 97.29 + 54.45[/tex]

        [tex]I= 312.09 \ kg \cdot m^2[/tex]

hi guys!!! i have no more points, can someone nice guess all of these for me? :)
1.What happens to the ocean water before the precipitation part of the water cycle
2.During which stage of the water cycle does water from the ocean form clouds?
3.what is a runoff??
4.Which statement about oceans is incorrect? A.Evaporation occurs when water is warmed by the sun. B.Most evaporation and precipitation occur over the ocean. C.97 percent of Earth's water is fresh water from the ocean. D.Water leaves the ocean by the process of evaporation
5.How does most ocean water return to the ocean in the water cycle

tysm to u who answers :)

Answers

1. The ocean water collects back in the ocean.

2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.

3. an excessive amount of water flowing from downslope along earths surface

4. A.Evaporation occurs when water is warmed by the sun.

5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.

Hope this answer correct ✌️

an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas

Answers

Answer:

Explanation:

Charge on an electron (q) = 1.6 * 10 ^ -19 C

Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec

We know that, Force exerted on moving particle moving through a magnetic field :

[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]

1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B

B =  0.08573 T

If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.

Answers

Answer:

D. A convex lens in air

Explanation:

This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens

The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is

Answers

Answer:

E = -8.23 ​​10⁻¹⁷ N / C

Explanation:

In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is

         E = k q₁ q₂ / r²

in this case

q₁ is the charge of the proton and q₂ the charge of the electron

         E = - k e² / a₀²

let's calculate

         E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²

         E = -8.23 ​​10⁻¹⁷ N / C

Question 2.
In the US, lengths are often measured in inches, feet, yards and miles. Let's do
some conversions. The definition of the inch is: 1 inch = 25.4 mm, exactly. A foot is
12 inches and a mile is 5280 ft, exactly. A centimetre is exactly 0.01 m or 10 mm.
Sammy is 5 feet and 5.3 inches tall.
a). What is Sammy's height in Inches? (answer to 3 significant figures)
(3)
b). What is Sammy's height in Feet? (answer to 3 significant figures)
what is Sammy's hight in feet according to this statement ​

Answers

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]

So,

[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.

Answers

Answer:

a. to the west.

Explanation:

An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground

Answers

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.

Answers

Answer:

137.69°

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Phase angle:

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

Learn more about RLC circuit:

https://brainly.com/question/372577?referrer=searchResults

Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.

Answers

Answer:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.

Answers

Answer:

I know the answer

Explanation:

We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.

You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.

Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.

So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.

Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).

Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth

Answers

Answer:

Vrel= 0.75c

Explanation:

See attached file

A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.

Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.

With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.

Answers

Answer:

(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV

(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Explanation:

Given;

radius of the circular loop, r = 31.0 cm = 0.31 m

initial magnetic field, B₁ = 0.7 T

final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T

duration of change in the field, t = 29

(a) The magnitude of induced emf in the loop while the magnetic field is increasing.

[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]

[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]

Where;

A is the area of the circular loop

A = πr²

A = π(0.31)² = 0.302 m²

[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]

(b) the magnitude of the induced voltage at a constant magnetic field

E = A x B/t

E = (0.302 x 1.61) / 3.9

E = 0.1247 V

E = 124.7 mV

Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV

A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x axis at 37 m/s in the positive direction. The second, with mass 22 g , moves along the y axis at 34 m/s in the positive direction. Find the velocity of third piece.

Answers

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?

a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100

Answers

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55

Answers

Answer:

C $0.75 my friend I wish it is right answer

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