The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?

Answer:

From an average distance of 886 million miles (1.4 billion kilometers), Saturn is 9.5 astronomical units away from the Sun. One astronomical unit (abbreviated as AU), is the distance from the Sun to Earth. From this distance, it takes sunlight 80 minutes to travel from the Sun to Saturn.

we have that from the Question"The average mean distance of Saturn from the sun is" it can be said that  Tthe average mean distance of Saturn from the sun is

From the Question we are told

The average mean distance of Saturn from the sun is

Generally

The Sun is the star of the milky way galaxy and its distance from every planet in the milky way determines in one way or another its properties and in-habitability

Saturn being a Planet of the milky way we see that Saturn is a significant distance away from sun

A distance of 1427 x 10^6 km or 886 696 691 miles

Therefore

The average mean distance of Saturn from the sun is

A distance of 1427 x 10^6 km or 886 696 691 miles

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Answer:

The answer should be D

Explanation:

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

Answer:

a)    σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] ,  b)  σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂,  σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]

Explanation:

a) The very useful concept of charge density is defined by

          σ = Q / A

In this case we have a circular disk

The are of a circle is

         A = π r²

in this case we have a hole in the center of radius r = b, so

         A_net = π r² - π r_ {hollow} ²

         A_ {net} = π (a² - b²)

whereby the density is

          σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]

b) The density of the other disk is

          σ = Q₂ / A₂

          σ = [tex]\frac{Q_2}{d^2}[/tex]

c) The total waxed load is requested by the larger circle

           Q_ {total} = Q₁ + Q₂

the net charge density, in the whole system is

          σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]

the area  is

          A_{total} = π a²

since the other circle is inside, we are ignoring the space between the two circles

          σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]

Answer:

Generally, the lowest overtone for a pipe open at one end and closed would be at  y / 4  where y represents lambda, the wavelength.

Since F (frequency) = c / y       Speed/wavelength

F2 / F1 = y1 / y2      because c is the same in both cases

F2 = y1/y2 * F1

F2 = 3 F1 = 750 /sec

Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe

and y1 = 3 y2

The second harmonic will be three times the first harmonic. The answer is 750 Hz

Closed pipes have odd multiples of frequencies or harmonics. That is,

If  [tex]F_{0}[/tex] = fundamental frequency = first harmonic

[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic

[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic

[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic

Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.

By using the formula above,

second harmonic will be 3 x 250 = 750Hz

Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz

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Answer:

0.375

Explanation:

For incompressible flow, we know that;

ρ1•v1•A1 = ρ2•v2•A2

Where;

ρ1 = density of fluid at position A

v1 = speed of fluid at position A

A1 = area of tube

ρ2 = density of fluid at position B

v2 = speed of fluid at position B

A2 = area of tube

We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.

Thus;

ρ1/ρ2 = (v2•A2)/(v1•A1)

Now, the tube will have the same height.

But we are given;

diameter of A = 12.00 cm = 0.12 m

diameter of B = 6 cm = 0.06 m

Thus;

A1 = π(d²/4)h = πh(0.12²/4)

A2 = πh(0.06²/4)

We are also given;

v1 = 12 m/s

v2 = 18 m/s

Thus;

ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))

πh/4 will cancel out to give;

ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)

ρ1/ρ2 = 0.375

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t[tex]_{min[/tex] =  ( m + 1/2) × λ/2n

we substitute

t[tex]_{min[/tex] =  ( 0 + 1/2) × 711 /2(1.21)

t[tex]_{min[/tex] = 0.5 × 711/2.42

t[tex]_{min[/tex] = 0.5 × 293.80165

t[tex]_{min[/tex] = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

This question is incomplete, the complete question is;

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).

Answer:

the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

V[tex]_B[/tex] = 54 km/hr

V[tex]_A[/tex] = 72 km/hr

α = 100 m

now, angular velocity of Bxy will be;

ω[tex]_B[/tex] = V[tex]_B[/tex] / α

so, we substitute

ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100

ω[tex]_B[/tex] = 15 / 100

ω[tex]_B[/tex] = 0.15 rad/s

Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Answer:

oil heats faster

Explanation:

Answer:

the temperature of the intermediate reservoir is 624.5 K

Explanation:

Given the data in the question  

The two Carnot heat engines are operating in series;

[ T[tex]_H[/tex] ]

  ↓

((1)) ⇒ W[tex]_{out[/tex]

  ↓

[ T[tex]_M[/tex] ]

   ↓

 ((2)) ⇒ W[tex]_{out[/tex]

[ T[tex]_L[/tex] ]

The maximum possible efficiency for any heat engine is the Carnot efficiency;

η[tex]_{rev[/tex] = 1 - [tex]\frac{T_L}{T_H}[/tex]

the thermal efficiencies if both engines are the same will be;

η[tex]_A[/tex] = η[tex]_B[/tex]

1 -  [tex]\frac{T_M}{T_H}[/tex] = 1 - [tex]\frac{T_L}{T_M}[/tex]

1 - 1 -  [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]

-  [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]

[tex]\frac{T_M}{T_H}[/tex] =  [tex]\frac{T_L}{T_M}[/tex]

T[tex]_M[/tex]² = T[tex]_L[/tex] × T[tex]_H[/tex]

T[tex]_M[/tex] = √(T[tex]_L[/tex] × T[tex]_H[/tex])

source temperature of the first engine T[tex]_H[/tex] = 1300 K

sink temperature of the second engine T[tex]_L[/tex] = 300 K

we substitute

T[tex]_M[/tex] = √(300 × 1300)

T[tex]_M[/tex] = √390000

T[tex]_M[/tex] = 624.4998 K ≈ 624.5 K

Therefore, the temperature of the intermediate reservoir is 624.5 K

1) Does a 1 kg object weight 9.8 newtons on the moon? why?

No. 1kg of mass does not weigh 9.8N on the moon.

Weight = (mass) x (gravity).

Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.

2) How much does a 3-kg object weigh (on earth) in newtons?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (3 kg) x (9.8 m/s² )

Weight = 29.4 N

3) How much does a 20-kg object weigh (on earth) in newton?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (20 kg) x (9.8 m/s² )

Weight = 196 N

4) What must happen for the mass of an object to change?

When an object moves, its mass increases.  The faster it moves, the greater its mass gets.  But this is all part of Einstein's "Relativity".  The object has to move at a significant fraction of the speed of light before any change can be noticed or measured.  So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.

5) What are 2 ways the weight of an object can change?

First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.

But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood.  So the weight can change even though the mass doesn't.

The weight of an object changes if you take it to a place where gravity is stronger or weaker.

Let's say we have an object whose mass is 90.72 kilograms.  Like me !    

As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons  (200 pounds).

. . . Fly me to the moon. Gravity = 1.62 m/s²  Weight = 147 Newtons (33 lbs)

. . . Drag me to Jupiter.  Gravity = 24.8 m/s²  Weight = 2,249 N (506 pounds)

My mass never changed, but my weight sure did.

Answer:

a) space only    t = 1.28 s

b) space+ atmosphere   t_ {total} = 1.28000003 s

Explanation:

The speed of light in each material medium is constant, which is why we can use the uniform motion relations

           v= x / t

a) let's look for time when it only travels through space

          t = x / c

          t = 3.84 10⁸/3 10⁸

          t = 1.28 s

b) we look for time when it travels part in space and part in the atmosphere

space

as it indicates that the atmosphere has a thickness of e = 30 10³ m

           t₁ = (D-e) / c

           t₁ = (3.84 10⁸ - 30.0 10³) / 3 10⁸

           t₁ = 1.2799 s

atmosphere

we use the refractive index

           n = c / v

           v = c / n

we substitute in the equation of time

           t₂ = e n / c

           t₂ = 30 10³   1,000293 /3 10⁸

           t₂ = 1.000293 10⁻⁴ s

therefore the total travel time is

           t_ {total} = t₁ + t₂

           t_ {total} = 1.2799+ 1.000293 10⁻⁴

           t_ {total} = 1.28000003 s

we can see that the time increase due to the atmosphere is very small

Answer:

Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.

The answer is Motion

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

Answer:

fem = 7.58 10⁻⁵ V

Explanation:

For this exercise we use Faraday's law

          fem =   [tex]- \frac{d \Phi _B}{dt}[/tex]

the magnetic flux is

          Ф_B = B. A = B A cos θ

Tje bold are vectros.  Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1

The red blood cell area is

          A =π r²

indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m

the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T,  therefore we can write it

          B = B₀ sin (wt) = B₀ sin( 2π f t)

we substitute

           fem = - A dB / dt

           fem = - A B₀  [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]

           fem = - π r² Bo (2πf  cos 2πft)

the maximum electromotive force occurs when the function is ±1

           fem = 2 π² r² B₀ f

let's calculate

           fem = 2π²  (8.00 10⁻³)²  1.00 10⁻³ 60

           fem = 7.58 10⁻⁵ V

Answer:

141.18 ohms

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12

Current (I) = 0.085 A

Resistance (R) =?

The resistance needed can be obtained as follow:

V = IR

12 = 0.085 × R

Divide both side by 0.085

R = 12 / 0.085

R = 141.18 ohms

Therefore, a resistor of resistance 141.18 ohms is needed.

Answer:

(a) 8.85×10⁻³ minutes

(b) 4.24×10¹⁹ electrons

Explanation:

(a) Using,

Q = it............................. Equation 1

Where Q = quantity of charge, i = current, t = time.

Make t the subject of the equation

t = Q/i............................. Equation 2

Given: Q = 6.8×10⁰ C, i = 12.8 A

Substitute these values into equation 2

t = 6.8×10⁰/12.8

t = 8.85×10⁻³ minutes

(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3

Where n = number of electrons.

Given: Q = 6.8×10⁰ C

Substitute into equation 2

n = 6.8×10⁰/1.602×10⁻¹⁹

n = 4.24×10¹⁹ electrons

(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b) Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

As we know Q = IT

Where Q = quantity of charge, i = current, T = time.

From the above equation

                    T= Q/I.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A

Substitute these values  

T=  [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8

T =  [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes

Now the number of the electrons present in the charge will be

n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])

Where n = number of electrons.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C

Substitute Value of Q  

n =  [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]

n = [tex]4.24\times\d10^{19}[/tex] electrons

Thus

(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b)Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

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Answer:

the amplitude of the wave

the energy of the wave

the type of wave

the type of medium

Answer:

0.08 ms^-2

Explanation:

by using v= u + at

initial velocity is zero as it is starting from rest

4= 0 + a x 50

4/50 = a = 0.08 ms^-2

Answer:

a. 3.81

Explanation:

F = GMm/r^2

F = (6.67 x 10^-11 x 28 x 1) / 7^2

F = 3.81 x 10^-11 N

Answer:

Force is an external agency that changes or tends to change the state of body from rest to motion or motion to rest.

The SI unit of force is newton(N)

Answer:

Explanation:

you can say the law of superpoition can tell us that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock or fossil in the rock is older if it is farther down in the rock layers. hope it helps

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

Answer:

533.33 nm

Explanation:

Since dsinθ = mλ  for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.

Since the fringes coincide,

m'λ = m"λ'

λ' = m'λ/m"

= 10 × 640 nm/12

= 6400 nm/12

= 533.33 nm

Answer:

50 protons 50 electrons and 69 neutrons...

Explanation:

the number of protons is equal to number of electrons. then mass number is the sum of protons and neutrons in a nucleus so for we to get the number of neutrons we take the mass number subtract the protons number.

Answer:

x-Speed/velocity

y-time.

Explanation:

because Speed is a rate of change of distance while time how long it takes a a car to move to a specific point

Answer:

static force

Explanation:

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