A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper plate and off the lower plate at a rate of 5 A. Find the time rate of change of the electric field between the plates.

Explanation:

It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.

As they are in equilibrium, the force of gravity due to each other is same. So,

[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]

So, the ratio of masses X/Y is 0.0847

Answer:

The  kinetic energy is [tex]KE = 7.59 *10^{10} \ J[/tex]

Explanation:

From the question we are told that

       The  radius of the orbit is  [tex]r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m[/tex]

       The gravitational force is  [tex]F_g = 6600 \ N[/tex]

The kinetic energy of the satellite is mathematically represented as

       [tex]KE = \frac{1}{2} * mv^2[/tex]

where v is the speed of the satellite which is mathematically represented as

     [tex]v = \sqrt{\frac{G M}{r^2} }[/tex]

=>  [tex]v^2 = \frac{GM }{r}[/tex]

substituting this into the equation

      [tex]KE = \frac{ 1}{2} *\frac{GMm}{r}[/tex]

Now the gravitational force of the planet is mathematically represented as

      [tex]F_g = \frac{GMm}{r^2}[/tex]

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     [tex]KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r[/tex]

=>    [tex]KE = \frac{ 1}{2} *F_g * r[/tex]

substituting values

       [tex]KE = \frac{ 1}{2} *6600 * 2.3*10^{7}[/tex]

         [tex]KE = 7.59 *10^{10} \ J[/tex]

Answer:

There are 586maxima

Explanation:

Pls see attached file

Answer:

a

  [tex]\phi = 1.78 *10^{-7} \ Weber[/tex]

b

 [tex]L = 1.183 *10^{-7} \ H[/tex]

Explanation:

From the question we are told that

   The radius is  [tex]r = 6 \ cm = \frac{6}{100} = 0.06 \ m[/tex]

   The current it carries is  [tex]I = 1.50 \ A[/tex]

The  magnetic flux of the coil is mathematically represented as

       [tex]\phi = B * A[/tex]

Where  B is the  magnetic field which is mathematically represented as

         [tex]B = \frac{\mu_o * I}{2 * r}[/tex]

Where  [tex]\mu_o[/tex] is the magnetic field with a constant value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

substituting  value

          [tex]B = \frac{4\pi * 10^{-7} * 1.50 }{2 * 0.06}[/tex]

          [tex]B = 1.571 *10^{-5} \ T[/tex]

The area A is mathematically evaluated as

       [tex]A = \pi r ^2[/tex]

substituting values

       [tex]A = 3.142 * (0.06)^2[/tex]

       [tex]A = 0.0113 m^2[/tex]

the magnetic flux is mathematically evaluated as    

        [tex]\phi = 1.571 *10^{-5} * 0.0113[/tex]

         [tex]\phi = 1.78 *10^{-7} \ Weber[/tex]

The self-inductance is evaluated as

       [tex]L = \frac{\phi }{I}[/tex]

substituting values

        [tex]L = \frac{1.78 *10^{-7} }{1.50 }[/tex]

         [tex]L = 1.183 *10^{-7} \ H[/tex]

..........................................................

The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².

Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.

Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.

E.g: Black-body radiation

Given the values are,

Wattage of bulb = W = 25 W

distance = d = 6.5 cm = 0.065 m

To know the Radiation Pressure,

It can be given by

P = I/c

Where, c = 299792458 m/s is the speed of light,

I is the intensity of radiation and given by

I = W/4πd²

Where W is the Wattage of bulb and d is the distance

I = 25/4π*0.065²

I = 470.872 w/m²

so, the radiation pressure becomes

P = 470.872/299792458

P = 1.5707x10⁻⁶ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²

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Answer:

Current that flows through any one of the resistors has a value of 12 amperes.

Explanation:

When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:

[tex]V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})[/tex]

[tex]i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}[/tex]

Where:

[tex]V_{batt}[/tex] - Voltage of the battery, measured in volts.

[tex]R_{1}[/tex], [tex]R_{2}[/tex], [tex]R_{3}[/tex] - Electric resistance of the first, second and third resistors, measured in ohms.

[tex]i[/tex] - Current, measured in amperes.

If [tex]R_{1} = R_{2} = R_{3} = R[/tex], then:

[tex]i = \frac{V_{batt}}{3\cdot R}[/tex]

Current that flows through any one of the resistors has a value of 12 amperes.

The current flows via any of the resistors should have a value of 12 amperes.

At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.

Here the circuit should be

vbatt = i.(R1 + R2+ R3)

i = Vbatt/R1 + R2 + R3

here

Vbatt means the voltage of the battery

R1,R2, and R3 means the resistance of the first, second and third resistors

I means the current

So, in the case when

R1 = R2 = R3 = R

So,

i = Vbatt/3.R

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Answer:

931.00ft-lb

Explanation:

Pls see attached file

The work done in moving the object from x = 1 ft to x = 18 ft is 935  ft-lb.

Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.

Given that force = 6x - 2 pounds.

So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]

= [ 3x² - 2x]¹⁸₁

= 3(18² - 1² ) - 2(18-1) ft-lb

= 935  ft-lb.

Hence, the work done is  935  ft-lb.

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Answer:

The double slit experiment showed for the first time that light can be interfered, producing bands of light and dark fringes on a screen. This phenomenon was a unique and typical characteristic of waves.

Explanation:

Th double slit experiment by Thomas Young proved, and sealed for the first time the wave nature of light; showing that light just as any other wave can produce interference which was a unique, typical phenomenon of waves. The Interference of light was shown by allowing light to pass through narrow slits and superimpose on a wall or screen, at a distance away from the slit, producing a clear pattern of light and dark fringes. This was the first experiment to proof that darkness can be produced by the addition of light on light. Interference is accompanied by a spatial redistribution of the optical intensity without violation of power conservation. The phenomenon of interference proved the intuitive ideas of Huygens regarding the wave nature of light against Newton's particle nature of light theory.

Answer:

the firtz agrees with the expression for the shape of the curve of diracion of a slit

Explanation:

The diffraction phenomenon is described by the expression

              a sin θ = m λ

where a is the width of the slit, t is the angle from the center of the slit, l is the wavelength and m is an integer that corresponds to the maximum diffraction.

the previous equation qualitatively describes the curve of the diffraction phenomenon the equation takes the form

             I = I₀ [(sin ππ a y / R λ) / π a y / Rλ]²

             I = I₀ ’[sin π a y /Rλ]²

             I₀ ’= I₀ / (π a y /Rλ)²

By reviewing the two expressions given

equation 1

 w sin θ = m λ

where w =a  w   is the slit width

we see that the firtz agrees with the expression for the shape of the curve of diracion of a slit

Equation 2

the squares are missing

Answer:

R₈₀ = 146.43 Ω

Explanation:

The resistance of a resistor depends upon many factors. One of the main factors of the change in resistance of a resistor is the change in temperature. The formula for the resistance at a temperature other than 20°C is given as follows:

R₈₀ = R₀(1 + αΔT)

where,

R₈₀ = Resistance of wire at 80°C = ?

R₀ = Resistance of wire at 20° C = 104 Ω

α = Temperature coefficient of resistance for copper = 0.0068 °C⁻¹

ΔT = T₂ - T₁ = 80°C - 20°C = 60°C

Therefore,

R₈₀ = (104 Ω)[1 + (0.0068°C⁻¹)(60°C)]

R₈₀ = 146.43 Ω

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.

[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.

Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:

[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]

And final speed is now calculated after clearing it:

[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]

[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Answer:

2.55m

Explanation:

Using 1/do+1/di= 1/f

di= (1/f-1/do)^-1

( 1/0.0500-1/0.0510)^-1

= 2.55m

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]

The induced emf in the shorter coil is calculated as;

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

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

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

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]

The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:

[tex]E=-\frac{\delta\phi}{\delta t}[/tex]

where E is the induced emf,

[tex]\phi[/tex] is the magnetic flux through the coil.

The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:

[tex]\phi = \frac{\mu_oNIA}{L}[/tex]

so;

[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]

where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.

[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]

The emf produced in the coil at the center of the solenoid which has 14 turns will be:

[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]

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Answer:

v = 73.75 m/s

Explanation:

It is given that,

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.

Let us assume we need to find the velocity of the rocket when it runs out of fuel.

Let v is the final speed. Using the third equation of kinematics as :

[tex]v^2-u^2=2as[/tex]

u = 0

[tex]v=\sqrt{2as} \\\\v=\sqrt{2\times 3.2\times 850}\\\\v=73.75\ m/s[/tex]

So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s

Explanation:

It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.

We need to find the speed before and after the ball bounce.

Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,

[tex]\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s[/tex]

Let v is the final speed when it bounces to a height of 1.5 m. So,

[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s[/tex]

So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.

Answer:

118.6 rad/s²

Explanation:

Δθ = 4100.0 rad

ω₀ = 159.0 rad/s

ω = 999.0 rad/s

Find: α

ω² = ω₀² + 2αΔθ

(999.0 rad/s)² = (159.0 rad/s)² + 2α (4100.0 rad)

α = 118.6 rad/s²

Answer:

I believe it is 91

Explanation:

Answer:

Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.

Explanation:

Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.

Answer:

D. Q/2

Explanation:

See attached file

Answer:

The critical angle is  [tex]i = 41.84 ^o[/tex]

Explanation:

From the  question we are told that

    The index of refraction of the sugar solution is  [tex]n_s = 1.5[/tex]

   The  index of refraction of air is  [tex]n_a = 1[/tex]

Generally from Snell's  law

      [tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]

Note that the angle of incidence in this case is equal to the critical angle

Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]

So  

      [tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]

      [tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]

      [tex]i = 41.84 ^o[/tex]

Answer:

The electric field strength between the plates can be increased by decreasing the length of each side of the plates.

Explanation:

The electric field strength is given by;

[tex]E = \frac{V}{d}[/tex]

where;

V is the electric potential of the two opposite charges

d is the distance between the two parallel plates

[tex]E =\frac{V}{d} = \frac{\sigma}{\epsilon _o} \\\\(\sigma = \frac{Q}{A} )\\\\E = \frac{Q}{A\epsilon_o} \\\\E = \frac{Q}{L^2\epsilon_o}[/tex]

Where;

ε₀ is permittivity of free space

L is the length of each side of the plates

From the equation above, the electric field strength can be increased by decreasing the length of each side of the plates.

Therefore, decreasing the length of each side of the plates, could be made to increase the strength of the electric field between the plates

Answer:

a

  [tex]F =326.7 \ N[/tex]

b

  [tex]M = 6[/tex]

Explanation:

From the question we are told that

          The mass of the rock is  [tex]m_r = 200 \ kg[/tex]

          The  length of the small object from the rock is  [tex]d = 2 \ m[/tex]

          The  length of the small object from the branch [tex]l = 12 \ m[/tex]

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      [tex]W = m_r * g[/tex]

substituting values

     [tex]W = 200 * 9.8[/tex]

     [tex]W = 1960 \ N[/tex]

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         [tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]

Here  [tex]\theta[/tex] is very small so  [tex]cos\theta * l = l[/tex]

                               and  [tex]cos\theta * d = d[/tex]

Hence

       [tex]F * l - W * d = 0[/tex]

=>    [tex]F = \frac{W * d}{l}[/tex]

substituting values

        [tex]F = \frac{1960 * 2}{12}[/tex]

       [tex]F =326.7 \ N[/tex]

The  mechanical advantage is mathematically evaluated as

          [tex]M = \frac{W}{F}[/tex]

substituting values

        [tex]M = \frac{1960}{326.7}[/tex]

       [tex]M = 6[/tex]

Answer:

To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:

2*(3u)*t = m* lambda

Making m the subject of the formula will give the formula:

m= 6*u*t/ Lambda

Given: Lambda= 633*10^9 while u and t are unknown

Therefore the value of m can be calculated given the formula below:

m= 6*u*t/ 633*10^9

Explanation:

To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:

2*(3u)*t = m* lambda

Making m the subject of the formula will give the formula:

m= 6*u*t/ Lambda

Given: Lambda= 633*10^9 while u and t are unknown

Therefore the value of m can be calculated given the formula below:

m= 6*u*t/ 633*10^9

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The spring scale [tex]F_2[/tex] reads  [tex]F_2 = 2.4225 \ N[/tex]

Explanation:

From the question we are told that

      The first force is  [tex]F_1 = 10.5 \ N[/tex]

      The acceleration by which the cart moves to the right is  [tex]a = 2.50 \ m/s^2[/tex]

      The mass of the cart is  m  = 3.231  kg

Generally the net force on the cart is  

       [tex]F_{net} = F_1 - F_2[/tex]

This net force is mathematically represented as

      [tex]F_{net} = m * a[/tex]

So  

        [tex]m* a = 10 - F_2[/tex]

        [tex]F_2 = 10.5 - 2.5 (3.231)[/tex]

        [tex]F_2 = 2.4225 \ N[/tex]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x  10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

[tex]F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N[/tex]

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

[tex]W = F_{net} *h\\\\W = 15.328 *10^{-3} * h[/tex]

W = K.E

[tex]15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m[/tex]

Therefore, the ball traveled 0.827 m

The height at which the ball goes for the given parameters is; 0.827 m

We are given;

distance between the metal plates; d = 3.1 m

mass of glass; m = 1.1g = 0.0011 kg

charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C

speed of the glass ball; v = 4.8 m/s

voltage of the ceiling; V = +3.0 × 10⁶ V

The repulsive force experienced by the ball is gotten from the formula;

F = qV/d

|F| = (4.7 × 10⁻⁹ × 3 ×  10⁶)/3.1

|F| = 4.548 × 10⁻³ N

F = -4.548 × 10⁻³ N (negative because it is repulsive force)

The net horizontal force experienced by the ball is;

F_net = F - mg

F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)

F_net = -15.328 × 10⁻³ N

To get the height of the ball, we will use the formula;

F_net * h = ¹/₂mv²

h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)

We took the absolute value of F_net, hence it is not negative

h = 0.827 m

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Answer:

Vx' = (Vx - u) / (1 - Vx *u / c^2)      velocity transformation formula

In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship

VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c

VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c

In both cases an observer on earth will observe the light traveling at speed c.

Answer:

The wait time is [tex]t_w = 3.4723 \ s[/tex]

Explanation:

From the question we are told that

    The distance of the hot air balloon above the ground is  [tex]z = 50 \ m[/tex]

    The distance of the hot air  balloon from the target is  [tex]k = 100 \ m[/tex]

    The  speed of the wind is  [tex]v = 15 \ m/s[/tex]

Generally the time it will take the balloon to hit the ground  is  

           [tex]t = \sqrt{ \frac{2 * z }{g} }[/tex]

where g is acceleration due to gravity with value [tex]g = 9.8 m/s^2[/tex]

   substituting values  

                  [tex]t = \sqrt{ \frac{2 * 50 }{9.8} }[/tex]

                 [tex]t = 3.194 \ s[/tex]

Now at the velocity the distance it will travel before it hit the ground is mathematically represented as

               [tex]d = v * t[/tex]

   substituting values

              [tex]d = 15 * 3.194[/tex]

             [tex]d = 47.916 \ m[/tex]

Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as  

         [tex]b = k - d[/tex]

   substituting values

        [tex]b =100 -47.916[/tex]

         [tex]b = 52.084 \ m[/tex]

Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as

        [tex]t_w = \frac{b}{v}[/tex]

substituting values

       [tex]t_w = \frac{52.084}{15}[/tex]

       [tex]t_w = 3.4723 \ s[/tex]

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

Answer:

a

    [tex]\alpha = 2327.7 \ rev/s^2[/tex]

b

   [tex]\theta = 9124.5 \ rev[/tex]

Explanation:

From the question we are told that

    The maximum  angular   speed is  [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]

     The  time  taken is  [tex]t = 2.8 \ s[/tex]

     The  minimum angular speed is  [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest

Apply the first equation of motion to solve for acceleration we have that

       [tex]w_{max} = w_{mini} + \alpha * t[/tex]

=>     [tex]\alpha = \frac{ w_{max}}{t}[/tex]

substituting values

       [tex]\alpha = \frac{40950.73}{2.8}[/tex]

       [tex]\alpha = 14625 .3 \ rad/s^2[/tex]

converting to [tex]rev/s^2[/tex]

  We have

           [tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]

           [tex]\alpha = 2327.7 \ rev/s^2[/tex]

According to the first equation of motion the angular displacement is  mathematically represented as

       [tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]

substituting values

      [tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]

      [tex]\theta = 57331.2 \ radian[/tex]

converting to revolutions  

        [tex]revolution = 57331.2 * 0.159155[/tex]

        [tex]\theta = 9124.5 \ rev[/tex]