Consider a neutron star with a mass equal to the sun, a radius of 19 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star

Answers

Answer 1

Answer:

120 km/s

Explanation:

Given data :

Radius of the star is r = 19 km

Rotational time period of the star is T = 1 s

Therefore, we know that the velocity of the star is given by :

[tex]$V=\frac{2\pi r}{T}$[/tex]

[tex]$V=\frac{2 \times 3.14 \times 19\times 10^3}{1}$[/tex]

V = 119380.52 m/s

Therefore, the velocity of the point on the equator of the star is = 120 km/s


Related Questions

Solve numerical problem. Please give me step - step explanation Help me out plz

Answers

Answer:

You should multiply 60 kg*9.8 and answer will come.

Hope this will help you.

Answer:

yes she is right you should multiple 60*9.8

have a great day God bless you

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.

Answers

Answer:

I think D) less than 50 years

Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.

Who is an astronaut?

An astronaut observes and performs the experiments based on the universe.

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.

Due to special relativity, between space and Earth, both moving with different speeds.

The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.

So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.

Thus, the correct option is d.

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a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection

Answers

Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.

Answers

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?

Answers

Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

1. A sequence of potential differences v is applied accross a wire (diameter =0.32 mm length = 11 cm and the resulting current I are measured as follows: V 0.1 0.2 0.3 0.4 0.5 I (MA) 72 144 216 288 360 2) a) plot a graph of v against I.
b) determine the wire's resistence , R.
c) State ohm's law and try to relate it . your results.​

Answers

Answer:

a. Find the graph in the attachment

b. 720 kΩ

c. The ratio V/I gives us our resistance which is 720 kΩ

Explanation:

a) plot a graph of V against I.

To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.

b) Determine the wire's resistance , R.

The resistance of the wire is determined as the gradient of the graph.

R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)

Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA

R = (144 V - 72 V)/(0.2 - 0.1) mA

R = 72 V/0.1 mA

R = 72 V/(0.1 × 10⁻³ A)

R = 720 × 10³ V/A

R = 720 kΩ

c) State ohm's law and try to relate it your results.​

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.

Mathematically, V ∝ I

V = kI

V/I = k = R

Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.

Since V/I is constant, we thus verify Ohm's law.

What is the need for satellite communication elaborate

Answers

The high frequency radio waves used for telecommunications links travel by line of sight and so are obstructed by the curve of the Earth. The purpose of communications satellites is to relay the signal around the curve of the Earth allowing communication between widely separated geographical points.

Explanation:

hope it helps!!

A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event

Answers

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

What about Iceland's location makes it particularly well-suited to produce electricity from geothermal energy

Answers

The volcanoes in Iceland generate geothermal energy, heating & help with the making of electricity

Answer:

Iceland lies on a boundary where two plates are moving away from each other. Heat from Earth’s interior rises through this plate boundary at a fast rate. This fact makes Iceland well-suited to producing electricity using its abundance of geothermal energy.

Explanation:

Edmentum sample answer.

15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?

Answers

Answer: (15 - 0)/1.8 = 8. 33m/s^2

Explanation:

The acceleration of the racehorse is 8.33 m/s²

The given parameters;

initial velocity of the racehorse, u = 0

final velocity of the racehorse, v = 15  m/s

time of motion of the horse, t = 1.8 s

The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;

[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]

Thus, the acceleration of the racehorse is 8.33 m/s²

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What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making

Answers

Hi! I believe the answer is C. or The seasons because all the other options use springs when making cars, bikes, and clocks. I hope this helped. Goodluck :)

Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for longer wavelengths of light. The work function of the surface is

Answers

Explanation:

Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]

[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]

[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]

The work function [tex]\phi[/tex] is then

[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]

[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]

The work function of the surface is equal to 3.98 × 10⁻¹⁹J.

What are frequency and wavelength?

The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.

The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.

The expression between wavelength (λ), frequency, and speed of light (c) is:

c = νλ

Given, the wavelength of the light, ν = 500 nm

The frequency of the light can determine from the above-mentioned relationship:

ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz

The work function = h ν =  6 × 10¹⁴ × 6.626 × 10⁻³⁴

φ = 3.98 × 10⁻¹⁹J

Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.

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Question 5 of 10
What must be the same for two resistors that are connected in parallel?

Answers

Answer:

in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .

Explanation:

Flapping flight is very energy intensive. A wind tunnel test
on an 89 g starling showed that the bird used 12 W of
metabolic power to fly at 11 m/s. What is its metabolic power for starting flight?

Answers

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]

Using the formula

Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction

Answers

Answer:

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Explanation:

The expression for the angle of refraction is

          n₁ sin θ₁ = n₂ sin θ₂

refractive index n₁ is for incident light and n₂ is for transmitted light.

We have two cases

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A. If they are separated by a distance 4.3 m, at what location between the wires is the net magnetic field be twice the strength of the magnetic field from wire 1

Answers

Answer:

x= 2*I1*d/(I1+I2) meter

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How far horizontally will it travel in 2 seconds?
A. 30 m
B. 90 m
C. 45 m
D. 60 m

Answers

Answer:

It will travel Vx * t = 30 m/s * 2 s = 60 m

Distance = velocity x time
= 30m/s x 2 sec
= 60 m/s

Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​

Answers

[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m

Answers

Answer:

[tex]v=(6ti+6k)\ m/s[/tex]

Explanation:

Given that,

The position of a particle is given by :

[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]

Let us assume we need to find its velocity.

We know that,

[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]

So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].

what Is accuracy ............​

Answers

Answer:

Accuracy is how much the consequence of an estimation adjusts to the right worth or a norm' and basically alludes to how close an estimation is to its concurred esteem

《OAmalaOHopeO》

Answer:

In a set of measurements, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.

Explanation:

_Hope it helps you_

In which situation should a parent be proactive and act to assume responsibility?

Answers

Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.

Explanation: i picked this and it is correct, you’re welcome:)

3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic

Answers

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

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Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.

Answers

D. I hope my answer helps you!

A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz

Answers

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

Given the following data:

Period = 645 μs

Note: μs represents microseconds.

Conversion:

1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds

645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

[tex]Frequency = \frac{1}{Period}[/tex]

Substituting the value into the formula, we have;

[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

Conversion:

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = [tex]0.001[/tex] × [tex]1550.39[/tex]

X = 155039 kHz

To 3 significant figures;

Frequency = 155 kHz

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

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Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]​

Answers

Answer:

Gap left = Change in length on heating

Gap=Initial length×Coefficient of linear expansion×change in temperature

Gap=10×0.000012×15m

⟹Gap=0.0018 m

this is an example u have to put your equation in it

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars

Answers

Answer:

the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Explanation:

Given the data in the question;

Time period = 6 months = 1.577 × 10⁷ s

orbital speed v = 64000 m/s

since its a circular orbit,

v = 2πr / T

we solve for r

r = vT/ 2π

r = ( 64000 × 1.577 × 10⁷ ) / 2π

r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU

Now, from Kepler's law

T² = r³ / ( m₁ + m₂ )

T = 6 months = 0.5 years

we substitute

(0.5)² = (1.0737)³ / ( m₁ + m₂ )

0.25 = 1.2378 / ( m₁ + m₂ )

( m₁ + m₂ ) = 1.2378 / 0.25

( m₁ + m₂ ) = 4.9512

m₁ = m₂  = 4.9512 / 2 = 2.4756 solar mass

we know that solar mass = 1.989 × 10³⁰ kg

so

m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg

m₁ = m₂ = 4.92 × 10³⁰ kg

Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2).
a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213

Answers

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

substituting the values of the variables into the equation, we have

k = mg/x  

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.

a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?

b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?

Answers

Answer: it would be daddy

Explanation:

Because I’m daddy

Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?

Answers

Answer:

The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.

Explanation:

Consider newtonian mechanics here.

Dynamic equation is

The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.

we'll consider newtonian mechanics here.

so the dynamic equations is S = ut + 0.5at^2

we know u=0.8

S=1.2+0.9

t=1.2/1.5

find a.

A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.

Answers

Answer:

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium

[tex]d_0[/tex] is the object distance

[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium

[tex]d_i[/tex] is the image distance

R is the radius of curvature

Now, let [tex]d_0[/tex] = ∞, such that;

[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

we make [tex]d_i[/tex] subject of the formula

[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )

[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )

given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00

so we substitute

[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

[tex]d_i[/tex] = 9.165 / 0.41

[tex]d_i[/tex] = 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm

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