g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria are prepared, each consisting initially of one bacterium. One culture is left on earth and the other placed on a rocket that travels at a speed of 0.893c relative to the earth. At a time when the earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to an earth-based observer

Answers

Answer 1

Answer:

86.4 hrs

Explanation:

The amount of bacteria is initially 1

It doubles every 24 hrs.

After first 24 hrs, the amount = 2

After next 24 hrs = 4

After next 24 hrs = 8

After next 24 hrs = 16

After next 24 hrs = 32

After next 24 hrs = 64

After next 24 hrs = 128

After next 24 hrs = 256

Total time taken to reach 256 = 24 x 8 = 192 hrs

For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship

t = t'(1 - ¥^2)^0.5

Where t is the contracted time =?

t' is the time on earth

¥ = v/c

Where v is the speed of the rocket

c is the speed of light

since v = 0.893c

¥ = 0.893

Substituting, we have

t = 192 x (1 - 0.893^2)^0.5

t = 192 x 0.2025^0.5

t = 192 x 0.45 = 86.4 hrs


Related Questions

A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glass.

Answers

Answer:

The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

[tex]v=\dfrac{c}{\mu}[/tex]

Put the value into the formula

[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]

[tex]v=1.06\times10^{8}\ m/s[/tex]

We need to calculate the time of short pulse of light beam

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]

[tex]t=2.37\times10^{-9}\ sec[/tex]

Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in a time T. The heat of fusion of this metal is

Answers

Answer:

L = Pt/M

Explanation:

Power, P= Q/t = mL/t

we know that, (Q=m×l)

Now ⇒l= Pt/M

Thus l= Pt/M

For a beam of light in air (n = 1) reflecting off glass (n = 1.5), what is Brewster's angle to the nearest degree?

Answers

Answer: 56°

Explanation:

Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.

The formula is;

[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]

[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]

= 56.30993247

= 56°

When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?

Answers

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

within which type of system is the total mass conserved but not the total energy

Answers

In a closed system the mass is conserved, but the energy is not conserved.

To find the answer, we have to study about different systems in thermodynamics.

What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.

Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.

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An electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Which of the following are also the same for the two particles?
(A) speed
(B) kinetic energy
(C) frequency
(D) momentum

Answers

Explanation:

The De-Broglie wavelength is given by :

[tex]\lambda=\dfrac{h}{p}[/tex]

h is Planck's constant

p is momentum

In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.

Only momentum is the factor that remains same for both particles i.e. momentum.

Kasek rides his bicycle down a 6.0° hill (incline is
6° with the horizontal) at a steady speed of 4.0
m/s. Assuming a total mass of 75 kg (bicycle and
Kasek), what must be Kasek's power output to
climb the same hill at the same speed? ​

Answers

Answer:

 P = 2923.89 W  

Explanation:

Power is

     P = F v

for which we must calculate the force, let's use Newton's second law, let's set a coordinate system with a flat parallel axis and the other axis (y) perpendicular to the plane

X Axis  

         F - Wₓ = 0

         F = Wₓ

Y Axis

         N -  [tex]W_{y}[/tex] = 0

let's use trigonometry for the components of the weight

         sin 6 = Wₓ / W

         cos 6 = W_{y} / W

         Wₓ = W sin 6

         W_{y} = W cos 6

          F = mg cos 6

          F = 75 9.8 cos 6

          F = 730.97 N

let's calculate the power

        P = F v

        P = 730.97 4.0

        P = 2923.89 W

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?

Answers

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      [tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

       

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?

Answers

Answer:

The separation of the two slits is 0.456 mm.

Explanation:

Given the wavelength of light = 519 nm

The indifference pattern = 4.6 m

Adjacent bright fringes = 5.2 mm

In the interference, the equation required is Y = mLR/d

Here, d sin theta = mL

L = wavelgnth

For bright bands, m is the  order = 1,2,3,4  

For dark bands,  m = 1.5, 2.5, 3.5, 4.5

R = Distance from slit to screen (The indifference pattern)

Y = Distance from central spot to the nth  order fringe or fringe width

Thus,  here d = mLR/Y

d = 1× 519nm × 4.6 / 5.2mm

d = 0.459 mm

A rigid container holds 4.00 mol of a monatomic ideal gas that has temperature 300 K. The initial pressure of the gas is 6.00 * 104 Pa. What is the pressure after 6000 J of heat energy is added to the gas?

Answers

Answer:

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Explanation:

When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.

In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:

[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]

Where:

[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.

If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:

[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]

[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]

And change in heat energy ([tex]Q[/tex]), measured by joules, by:

[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]

Where:

[tex]n[/tex] - Molar quantity, measured in moles.

The final temperature of the monoatomic ideal gas is now cleared:

[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]

Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:

[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]

[tex]T_{2} = 420.279\,K[/tex]

The final pressure of the system is calculated by the following relationship:

[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]

If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:

[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]

[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for the blood in an artery during heart surgery). Such a device is illustrated. The conducting fluid moves with velocity v in a tube of diameter d perpendicular to which is a magnetic field B. A voltage V is induced between opposite sides of the tube. Given B = 0.120 T, d = 1.2 cm., and a measured voltage of 2.88 mV, determine the speed of the blood.

Answers

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

[tex]E = Blv[/tex]

where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

[tex]v[/tex] is the velocity of the fluid through the field = ?

[tex]B[/tex] is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]

2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]

[tex]v[/tex] = 2.88/1.44 = 2 m/s

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?

Answers

Answer:

a

    [tex]z= 2.5 \ m[/tex]

b

   [tex]z = (1 \ m , 4 \ m )[/tex]

Explanation:

From the question we are told that

     Their distance apart is  [tex]d = 5.00 \ m[/tex]

      The  wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              [tex]z - (d-z) = m \lambda[/tex]

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        [tex]z - (5-z) = 0[/tex]

=>     [tex]2 z - 5 = 0[/tex]

=>     [tex]z= 2.5 \ m[/tex]

Generally the path difference for destructive  interference is

           [tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]

=>         [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]

=>        [tex]|2z - d| =\frac{\lambda}{2}[/tex]

substituting values

          [tex]|2z - 5| =\frac{6}{2}[/tex]

=>      [tex]z = \frac{5 \pm 3}{2}[/tex]

So  

      [tex]z = \frac{5 + 3}{2}[/tex]

      [tex]z = 4\ m[/tex]

and

      [tex]z = \frac{ 5 -3 }{2}[/tex]

=>   [tex]z = 1 \ m[/tex]

=>    [tex]z = (1 \ m , 4 \ m )[/tex]

A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.

Answers

Answer:

Explanation:

When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .

Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .

hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .

Suppose a 58-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.150 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.10 T? magnitude V direction ---Select--- †\

Answers

Answer:

95.7v

Explanation

Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

E= Ndစ/dt

E= N BA-0/ deta t

Given that

N = 58turns

B = 1.10T

A = 0.150m^²

Deta t= 0.1s

now we have

E = 58(1.10x0.150)/0.1

= 95.7v

Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE

Explanation:

A light beam is traveling through an unknown substance. When it strikes a boundary between that substance and the air (nair≈1), the angle of reflection is 29.0∘ and the angle of refraction is 39.0∘. What is the index of refraction n of the substance?

Answers

Answer:

0.7707

Explanation:

From Snell's law,

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. Thus this means that θ1 = 29°

=> 1.0003 * sin29 = n(2) * sin39

n(2) = (1.0003 * sin29) / sin39

n(2) = 0.7707

Explanation:

The index of refraction n of the substance is 0.7707

Snell law:

Here we know that

n(1) * sin θ1 = n(2) * sinθ2

here

n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection should be via the unknown substance that represent the same as the angle of incidence of air.

So,

θ1 = 29°

1.0003 * sin29 = n(2) * sin39

n(2) = (1.0003 * sin29) / sin39

n(2) = 0.7707

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A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary

Answers

Answer:

12A

Explanation:

Formula for calculating the relationship between  the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:

[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]  where;

Vs and Vp are the emf in the secondary and primary coil respectively

Ns and Np are the number if turns in the secondary and primary coil respectively

Ip and Is are the currents in the secondary and primary coil respectively

Since the are all equal to each other, then we can equate any teo of the expression as shown;

[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

Given parameters

Np = 500-turns

Ns = 2000-turns

Is = 3.0Amp

Required

Current in the primary coil (Ip)

Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]

[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]

Hence the current in the primary coil is 12Amp

Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine

Answers

It depends on what stage of the mission you're talking about.

==>  While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.

==>  When the engines ignite, their thrust (d) is greater than the force of gravity.  So the net force on the rocket is upward, and the spacecraft accelerates upward.

==>  After the engines shut down, the net force acting on the rocket is due to Gravity (c).

. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.  

. . . If it has enough horizontal speed, it enters Earth orbit.  

. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.    

A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.

What unbalanced force bills for the course of the internet pressure of the rocket?

A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.

What's the net pressure of unbalanced?

If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.

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15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current

Answers

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

      B = μ₀ N/L   I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

      B = (μ₀ I / L)  N

the amount between paracentesis constant, in the case of 4 loop the field is worth

      B = cte 4

N       B

4       4 cte

3       3 cte

2       2 cte

1        1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 31° from the edge of the building with an initial velocity of 15 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?

Answers

Answer:

69.58 m tall

Explanation:

Pls see attached file

In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit

Answers

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .

Answers

Answer:

364days

Explanation:

Pls see attached file

Explanation:

The moon will take 112.7 days to make one revolution around the planet.

What is Kepler's third law?

The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.

Given, the distance from the moon's center to the planet's surface,

h = 2.165 × 10⁵ km,

The radius of the planet, r = 4175 km  

The mass of the planet = 6.70 × 10²² kg

The total distance between the moon's center to the planet's center:

a = r +h = 2.165 × 10⁵ + 4175

a = 216500 + 4175

a = 220675

a = 2.26750 × 10⁸ m

The period of the planet can be calculated as:

[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]

[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]

T = 9738253.26 s

T = 112.7 days

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From a hot air balloon 2 km​ high, a person looks east and sees one town with angle of depression of 16 degrees. He then looks west to see another town with angle of depression of 84 degrees. What is the distance between the two towns?

Answers

Answer:

7km

Explanation:

The angle of depression is congruent to the angle of elevation and can be explained as angle below horizontal in which the person observing an object must view for him/her to view object's that are lower than him/her.

In angle of depression, there is assumption that object is closer to the person observing it, so there is parallel horizontal for both observing and object been observed.

hot air balloon 2 km​ high,

there exist two triangles

From trigonometry

Tanx= opposite/adjacent

Opp= 2km

Adj= X1

first triangle have base length of

Tan(16)=2/X1

X1=2/ tan(16)

X1=6.97

For Second triangle

Tanx= opposite/adjacent

Opp= 2km

Adj= X2

the other with a base length of

X2=2/tan(84)

X2=0.21

Therefore,, the total distance between the two towns is

x1+x2=6.97+0.21=7.18km

The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be

Answers

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

Answers

Answer:

[tex]y = 0.0394 \ m[/tex]

Explanation:

From the question we are told that

        The  distance of the screen is  [tex]D = 2.20 \ m[/tex]

       The distance of separation of the slit is  [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]

        The  wavelength of light is  [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]

Generally the condition for constructive interference is

            [tex]dsin\theta = n * \lambda[/tex]

=>        [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]

here n = 1 because we are considering the central diffraction peak

=>        [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]

=>       [tex]\theta = 1.0274 ^o[/tex]

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           [tex]y = D tan (\theta )[/tex]

substituting values

        [tex]y = 2.20 * tan (1.0274)[/tex]

        [tex]y = 0.0394 \ m[/tex]

You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination

Answers

Answer:

4 / 7

Explanation:

1/total resistance = 1/1 + 1/2 + 1/4

= 1¾

total resistance = 1 ÷ 1¾

= 4/7

Just wondering if I did this right

Answers

Yeah

All they are all correct

g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower

Answers

Answer:

The projectile strikes the tower at a height of 354.824 meters.

Explanation:

The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:

Horizontal motion

[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]

Vertical motion

[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.

[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

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

[tex]t[/tex] - Time, measured in seconds.

The time spent for the projectile to strike the tower is obtained from first equation:

[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]

If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:

[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]

[tex]t \approx 7.071\,s[/tex]

Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])

[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]

[tex]y \approx 354.824\,m[/tex]

The projectile strikes the tower at a height of 354.824 meters.

A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s

Answers

Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days?

Answers

Answer: 0.0617

Explanation:

Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15

Let X be a binomial variable that represents the number of days having wet weather.

Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.

Here, n= 7 ( 1 week = 7 days)

The probability that it will take a week for it three wet weather on 3 separate days:

[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]

Hence, the required probability =0.0617

Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is

Answers

Answer:

the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle

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