This problem concerns the properties of circular orbits for a satellite of mass m orbiting a planet of mass M in an almost circular orbit of radius r. In doing this problem, you are to assume that the planet has an atmosphere that causes a small drag due to air resistance. "Small" means that there is little change during each orbit so that the orbit remains nearly circular, but the radius can change slowly with time. The following questions will ask about the net effects of drag and gravity on the satellite's motion, under the assumption that the satellite's orbit stays nearly circular. Use G if necessary for the universal gravitational constant.

What is the potential energy U of the satellite?Express your answer in terms ofm, M, G, and r.What is the kinetic energy K of the satellite?Express the kinetic energy in termsof m, M, G, and r.

Answer:

The velocity of the ions relative to the earth observer is  [tex]v = 0.63 c[/tex]

Explanation:

From the question  we are told that

     The speed of the alien ship is  [tex]v_a = 0.3c[/tex]

     The speed of the iron gun is  [tex]v_g = 0.4c[/tex]

The motion of the ions  relative to the earths observer can be mathematically represented as

       [tex]v = \frac{v_a + v_g}{[\frac{1+ v_av_g}{c^2} ]}[/tex]

substituting values

     [tex]v = \frac{ 0.3c + 0.4c}{[\frac{1+ 0.3c * 0.4c}{c^2} ]}[/tex]

 =>   [tex]v = 0.63 c[/tex]

Answer:

a = ½ ρ A/M   v₁²

Explanation:

This is a problem of fluid mechanics, where the jet of water at constant speed collides with a paddle, in this collision the water remains at rest, we write the Bernoulli equation, we will use index 1 for the jet before the collision the index c2 for after the crash

           P₁ + ½ ρ v₁² + ρ g h₁ = P₂ + ½ ρ v₂² + ρ g h₂

in this case the water remains at rest after the shock, so v₂ = 0, as well as it goes horizontally h₁ = h₂

          P₁-P₂ = ½ ρ v₁²

         ΔP = ½ ρ v₁²

let's use the definition of pressure as a force on the area

         F / A = ½ ρ v₁²

         F = 1/2 ρ A v₁²

the density is

          ρ = m / V

the volume is

           V = A l

           F = ½ m / l v₁²

knowing the force we can focus on the acceleration of the mass palette M

         F = M a

         a = F / M

          a = ½ m/M  1/l   v₁²

as well it can be given depending on the density of the water

          a = ½ ρ A/M   v₁²

Answer:

a)  it is essential that the waves lurk coherently

b)the light passes through the slits, the relative phase between the two rays is due to the optical path difference of each one,

Explanation:

In double slit interference experiments it is essential that the waves lurk coherently, that is, that the relative phase of the waves that reaches each slit is maintained over time, this is achieved when point sources are used by passing light through a initial slit and if a laser is used it is already consistent at the output.

When the light passes through the slits, the relative phase between the two rays is due to the optical path difference of each one, when this path difference is equal to a whole number of donut lengths, it has constructive interference and if it is a number sowing we have destructive interference.

                   d sin θ = m λ                 constructive

                   d sin θ = (m + ½) λ         destructive

where d is the distance between the two slits, lam is the wavelength and m is an integer called the interference order

Answer:

1

Explanation:

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Answer:

If one places ones's hands into a sink full of hot water, heat is transferred from the hot water to the hands. Heat only flows when there is a temperature gradient, i.e when there is a difference in temperature between a hot and a cold body. Heat is always transferred from a hot body to a cold body and never the reverse in a normal system.

At the boundary of the hand and water, heat transfer is by conduction between the hot water molecules and the molecules of the hands. This heat is then further transferred to the internal body organs and body fluid through either convection or conduction. This heat transfer is maintained until both the water and the hands are at the same temperature, assuming the owner is able to withstand the heat.

Answer:[tex]3.6\ kW[/tex]

Explanation:

Given

Power Supplied [tex][tex]P_{input}=4\ kW[/tex][/tex]

Efficiency of the motor [tex]\neta =90\%[/tex]

and [tex]\neta =\dfrac{\text{Power output}}{\text{Power input}}[/tex]

[tex]\Rightarrow 0.9=\dfrac{P_{out}}{4}[/tex]

[tex]\Rightarrow P=0.9\times 4[/tex]

[tex]\Rightarrow P=3.6\ kW[/tex]

So, vacuum cleaner delivers a power of [tex]3.6\ kW[/tex]

Answer:

a = d > b = c

Explanation:

The information about amplitudes and phase differences for four pairs of waves of equal wavelengths are given below:

(a) 2 mm, 6 mm, and π rad

(b) 3 mm, 5 mm, and π rad

(c) 7 mm, 9 mm, and π rad

(d) 2 mm, 2 mm, and 0 rad

Whenever a wave has zero phase difference, its amplitude of the resultant wave will be twice the amplitude of any of the two waves. Nevertheless, let assume that the amplitude is a vector having angle Ø between them. The resultant vector will help us rank the four pairs according to the amplitude of their resultant wave by using phasor diagrams.

a.) 6 - 2 = 4mm

b.) 5 - 3 = 2mm

c.) 9 - 7 = 2mm

d.) 2 + 2 = 4mm

Therefore,

a = d and b = c

a = d > b = c

Please find the attached file for the phasor diagrams

Answer:

a)y = 485 m ,  v = 220 m / s , b)  y = 2954.39 m , c)   t_total = 51 s ,

d) v = 240.59 m / s

Explanation:

a) We can use vertical launch ratios for this exercise

the speed of the rocket the run out the fuel is

        v = v₀ + a t

the rocket departs with initial velocity v₀ = 0

        v = a t

        v = 55 4

        v = 220 m / s

the height at this point is

        y = y₀ + v₀t + ½ a t²

        y = y₀ + 1/2 a t²

        y = 45 + ½ 55 4²

        y = 485 m

b) the maximum height of the rocket is when its speed is zero

for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m

        v² = v₀´² + 2 g (y-y₀´ )

         0 = v₀´² + 2 g (y-y₀´ )

         y = y₀´ + v₀´² / 2g

         y = 485 + 220 2/2 9.8

         y = 2954.39 m

c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)

let's calculate the rise time

           v = v₀´- g t

           v = 0

            t₂ = v₀´ / g

            t₂ = 220 / 9.8

            t₂ = 22.45 s

Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor

           y = y₀´´ + v₀´´ t - ½ g t²

           0 = y₀´´ - ½ g t²

          t = √ (2 y₀´´ / g)

          t = √ (2 2954.39 / 9.8)

          t = 24.55 s

the total flight time is

       t_total = t₁ + t₂ + t₃

       t_total = 4 + 22.45 + 24.55

        t_total = 51 s

d) the veloicda right now

       v = vo + g t

       v = 9.8 24.55

       v = 240.59 m / s

Answer:

(b) It is located the right distance from its star to enable liquid water to exist on its surface.

Explanation:

The habitable zone according to astronomy and astrobiology is a region around a star that the planets around the star can hold and support liquid water. For our solar system, the habitable zone coincides approximately with the distance from the sun to Earth.

Answer:

b

Explanation:

If a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent then the chemical properties of the solute and solvent are different, therefore the correct answer is option B.

A chemical compound is a combination of two or more either similar or dissimilar chemical elements.

For example, H₂O is a chemical compound made up of two oxygen atoms and a single hydrogen atom.

As given in the problem a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent,

The correct response is option B because the chemical characteristics of the solute and solvent are different if the material is repeatedly boiled and stirred without the solute and solvent mixing.

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Answer:

Kinetic energy is the energy of motion

Answer:

Energy which a body possesses by virtue of being in motion

Explanation:

Answer:

Time taken for trip = 12.74 hour (Approx)

Explanation:

Given:

Distance of trip = 710-mi

Average speed for the trip = 55.7 mi/h

Find:

Time taken for trip = ?

Computation:

⇒ Time = Distance / Speed

⇒ Time taken for trip = Distance of trip / Average speed for the trip

⇒ Time taken for trip = 710-mi / 55.7 mi/h

⇒ Time taken for trip = 12.74 hour (Approx)

Answer:

E = 0.5N/C

Explanation:

In order to calculate the magnitude of the electric field you use the following formula:

[tex]E=\frac{F}{q}[/tex]

q: charge = 5.0*10^-4 C

F: force on the charge = 2.5*10^-4N

You replace the values of q and F in the equation for E:

[tex]E=\frac{2.5*10^{-4}N}{5.0*10^{-4}C}=0.5\frac{N}{C}[/tex]

hence, the magnitude of the electric field at the position of the carge is 0.5N/C

At the location of the test charge, the magnitude will be:

"0.5 N/C".

Whenever charge seems to be available inside any form, an electric property has been linked among each spatial position. This same size and direction are represented either by the quantity of E.

According to the question,

Charge, q = 5.0 × 10⁻⁴ C

Force on the charge, F = 2.5 × 10⁻⁴ N

We know the relation of the magnitude of electric field be

→ E = [tex]\frac{F}{q}[/tex]

By substituting the values, we get

     = [tex]\frac{2.5\times 10^{-4}}{5.0\times 10^{-4}}[/tex]

     = 0.5 N/C

Thus the approach above is correct.

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Answer:

It may turn into sedimentary rock when exposed to pressure.

Explanation:

When successive layer of sand is deposited on top of sand, the bottom layers go deeper and deeper. They are exposed to high pressures and temperatures and turn into hard rocks.

Answer:

bro i think ur from the uni of arid agriculture rawalpindi and today is your networking paper

Explanation:

Given:

To be calculated:

Calculate the acceleration of given object ?

Formula used:

Acceleration = v - u / t

Solution:

We know that,

Acceleration = v - u / t

☆ Substituting the values in the above formula,we get

Acceleration ⇒ 300 - 200 / 20

⇒ 100/20

⇒ 5 m/s²

A. All living things and objects

Answer:

A: All living things and objects. I Hope This Helped!

Explanation:

Describes Matter.

Answer:

62.4m

Explanation:

a) Horizontal distance traveled = x = [tex]v_x[/tex] * t

where,

[tex]v_x[/tex]  = horizontal velocity

and t = time in the air)

Time in the air t can be solved using the equation for y:

[tex]i_y=y_o+v_o_yt - 0.5gt^2[/tex]

where, [tex]y_o[/tex] = initial height i.e 0.85 m

[tex]v_o_y[/tex] = initial vertical velocity

and g = 9.8 m/s^2

When y = 0, the dog has hit the water.

So set y = 0 and solve for t.  

[tex]v_o_y[/tex] = 8.62 m/s [tex]\times[/tex] (sin 28) = 4.04 m/s

0 = 0.85 m + (4.04 m/s)t - 0.5gt²  

0.5t²-4.04t-0.85  

Solve this quadratic formula for t: the solutions are t = 8.2 s and t = -0.2. Reject the negative solution, so t = 8.2 s.

How far does the dog travel horizontally in 9.2 s?

x = [tex]v_x[/tex] * t  = (8.62 m/s) [tex]\times[/tex] (cos 28) [tex]\times[/tex]  8.2 s = 62.4 m  

She travels the horizontal distance before hitting the surface of the water is 62.4m

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion.

a) Horizontal distance traveled = x = V(x) * t

where, V(x)  is the horizontal velocity and t = time in the air

Time in the air t can be solved using the equation for y:

hy =yo + Voy x t - 1/2gt²

where,  yo= initial height i.e 0.85 m and Voy = initial vertical velocity

and g = 9.8 m/s^2

When y = 0, the dock has hit the water.

So set y = 0 and solve for t.  

Voy = 8.62 m/s x (sin 28) = 4.04 m/s

Substitute the values, we get time as

0 = 0.85 m + (4.04 m/s)t - 0.5gt²  

0.5t²-4.04t-0.85 =0

Solving the quadratic formula for t, we have

time t = 8.2 s and t = -0.2.

As time can't be negative, so, t = 8.2 s.

The horizontal distance is

x = Vo * t  = (8.62 m/s)  (cos 28)   8.2 s = 62.4 m  

Thus, she travels 62.4 m.

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Answer:

a) 342 m/s

b) 51*10^-6 m/s

c) 0.87m/s^2

Explanation:

The following function describes the displacement of the molecules in a sound wave:

[tex]s(x,t)=3.00nm\ cos(50.00\ m^{-1}x-1.71*10^4s^{-1}t)[/tex]  (1)

The general form of a function that describes the same situation is:

[tex]s(x,t)=Acos(kx-\omega t)[/tex]   (2)

By comparing equations (1) and (2) you have:

k: wave number = 50.00 m^-1

w: angular frequency = 1.71*10^4 s^-1

A: amplitude of the oscillation = 3.00nm

a) The speed of the sound is obtained by using the formula:

[tex]v=\frac{\omega}{k}=\frac{1.71*10^4s^-1}{50.00m^{-1}}=342\frac{m}{s}[/tex]

b) The maximum speed of the molecules is the maximum value of the derivative of s(x,t), in time. Then, you first obtain the derivative:

[tex]\frac{ds}{st}=-\omega A sin(kx-\omega t)[/tex]

The max value is:

[tex]v_{max}=\omega A[/tex]

[tex]v_{max}=(1.71*10^4s^-1)(3.00nm)=51300\frac{nm}{s}=51\frac{\mu m}{s}[/tex] = 51*10^-6 m/s

c) The acceleration is the max value of the derivative of the speed, that is, the second derivative of the displacement s(x,t):

[tex]a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=-\omega^2A cos(kx-\omega t)\\\\a_{max}=\omega^2 A[/tex]

Then, the maximum acceleration is:

[tex]a_{max}=(1.71*10^4s^{-1})^2(3.00nm)=0.87\frac{m}{s^2}[/tex]

Answer:

[tex]a)-3346.8\;N \\ b)-4937.04\;N-m[/tex]

Explanation:

a) - In Free-body diagram :

At point D, the free body diagram of a man :

[tex]D = Fn\\Fn=T+mg\\put\;values\;in\;it\\ .\;\;\;\;=300+(80)(9.81)=1084.8\;N[/tex]

[tex]Mg=200\times9.81=1962\;N[/tex]

[tex]\sum Fx=0\; where\; Ax=0[/tex]

[tex]\sum Fy=0\; where\; Ay-Mg-Fn-T=0[/tex]

Then, put the value in the equation.

[tex]Ay=3346.8\;N[/tex]

b)-

[tex]Ma=Mg(AE)+Fn(AD)+T=4937.04\;N-m[/tex]

Answer:

The phenomenon of beats is the result of sound interference and sound diffraction in periodic vibrations. In periodic vibration, the beats of sounds produced by the interference and diffraction of the waves of different frequencies.

Explanation:

Beats. When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats.

Expand each vector into their component forms:

[tex]\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N[/tex]

Similarly,

[tex]\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N[/tex]

[tex]\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N[/tex]

Then assuming the resultant vector [tex]\vec R[/tex] is the sum of these three vectors, we have

[tex]\vec R=\vec A+\vec B+\vec C[/tex]

[tex]\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N[/tex]

and so [tex]\vec R[/tex] has magnitude

[tex]\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N[/tex]

and direction [tex]\theta_R[/tex] such that

[tex]\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ[/tex]

Answer:

It will take the wheel 278.9 s to come to a stop

Explanation:

Mass of the potter's wheel, M = 30 kg

Diameter of the potter's wheel, d₁ = 60 cm = 0.6 m

Radius, r₁ = d/2 = 0.6/2

r₁ = 0.3 m

The moment of inertia of the wheel, [tex]I = 0.5Mr_1^{2}[/tex]

[tex]I = 0.5*30*0.3^{2}\\I = 1.35 kg.m^2[/tex]

d₂ = 14 cm = 0.14 m

r₂ = 0.14/2 = 0.07 m

Angular velocity, [tex]\omega = 180 rpm[/tex]

[tex]\omega = \frac{180*2\pi }{60} \\\omega = 18.85 rad/s[/tex]

Frictional Force, F = 1.3 N

The torque generated:

[tex]\tau = F*r_{2}\\\tau = 1.3*0.07\tau = 0.091 Nm[/tex]

Torque can also be calculated as:

[tex]\tau = I \alpha\\\tau = I \frac{\omega }{t} \\0.091 = 1.35*\frac{18.8 }{t} \\t = (18.8*1.35)/0.091\\t = 278.9 s[/tex]

The time taken for the potter's wheel to come to a stop is 280 s.

The given parameters;

The momentum of inertia of the potter's wheel is calculated ;

[tex]I = \frac{1}{2} Mr^2\\\\I = (0.5)(30)(0.3)^2\\\\I = 1.35 \ kgm^2[/tex]

The angular speed of the potter's wheel is calculated as follows;

[tex]\omega = 180 \ \frac{rev}{\min} \ \times \ \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 18.85 \ rad/s[/tex]

The time taken for the wheel to come to a stop is calculated as;

[tex]Fr = I \alpha \\\\Fr= I \times \frac{\omega}{t} \\\\t = \frac{I \omega }{Fr} \\\\[/tex]

d = 14 cm, r = 7 cm = 0.07 m

[tex]t = \frac{1.35 \times 18.85}{1.3 \times 0.07} \\\\t = 279.6 \ s\\\\t\approx 280 \ s[/tex]

Thus, the time taken for the potter's wheel to come to a stop is 280 s.

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Answer:

Explanation:

Given that:

Heater temperature ,T₁ = 1000K

Vaccum Chamber ,T₂ = 300K

emissivity of heater E₁ = 0.75

emissivity vaccum E₂ = 0.25

Heater diameter d₁ = 10 * 10⁻³mm

vaccum chamber d₂ = 50 * 10⁻³mm

When there is vaccum, then no air resistance will be there,

F₁₂ = 1

F₁₁ = 0

[tex]R_1= \frac{1-E_1}{E_1A_1} \\\\=\frac{1-0.75}{0.75*\pi * 10^-^2*L}[/tex]

[tex]R_2=\frac{1}{F_1_2 * A_1} \\\\=\frac{1}{1* \pi *10^-^2*L}[/tex]

[tex]R_3=\frac{1-0.25}{F_1_2 * A_1} \\\\=\frac{1}{0.25* \pi *5*10^-^2*L}[/tex]

Heat leaving from heater surface 1 to vaccum

[tex]Q_1_2 = \frac{L \pi \sigma (T_1^4- T_2^4)}{R_1+R_2+R_3}[/tex]

[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)}{\frac{0.25}{0.75*10^-2}+\frac{1}{10^-2} +\frac{0.75}{0.25*10^-^2*5} }[/tex]

[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)} {0.3333+1+0.6}\\\\Q_1_2= 91.39 \text {watt}[/tex]

Answer:radiation

Explanation:

radiation is the only one that makes sense

Answer:

WT =  3.32*10^34 J

Explanation:

The work done by the gravitational attraction between the Sun and the Earth in one complete orbit of the Earth can be calculated by using the following formula:

[tex]W_T=\int F_g dr[/tex]  (1)

Fg: gravitational force between Sun and Earth

The gravitational force is given by:

[tex]F_g=G\frac{m_sm_e}{r^2}[/tex]  (2)

G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2

ms: mass of the sun = 1.989*10^30 kg

me: mass of the Earth = 5.972 × 10^24 kg

r: distance between Earth and Sun, this value is a constant r = R = 149,597,870 km

You replace the formula (2) in (1):

[tex]W_T=\int G\frac{m_sm_e}{R^2}dr=G\frac{m_sm_e}{R^2}\int dr\\\\W_T=G\frac{m_sm_e}{R^2}(2\pi R)=2\pi G\frac{m_sm_e}{R}[/tex]

Next, you replace the values of all variables and solve obtain WT:

[tex]W_T=2\pi (6.674*10^{-11}m^3kg^{-1}s^{-2})\frac{(1.989*10^{30}kg)(5.972*10^{24}kg)}{(149597870*10^3 m)}\\\\W_T=3.32*10^{34}J[/tex]

hence, the work done on the Earth, in one orbit, is 3.32*10^34 J

Answer:

im sure your already past this but it's E.

Explanation:

This is because in this case potential energy is linear to height, which means that the higher the more potential energy.

Answer:

Explanation:

capacitance of the capacitor C = 4πε₀R

= 1 x .103 / (9 x 10⁹ )  [ 1/ 4πε₀  = 9 x 10⁹ . ]

= .0114 x 10⁻⁹ F

potential  V = 4.5 x 10³ v

charge Q = CV

.0114 x 10⁻⁹  x 4.5 x 10³

= .0515 x 10⁻⁶ C

charge on one electron = 1.6 x 10⁻¹⁹

no of electrons to be removed

= .0515 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .032 x 10¹³

3.2 x 10¹¹ electrons.