(a) What is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 J of useful work while metabolizing 500 kcal of food energy? % (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%? kcal

Answers

Answer 1

Answer:

a) The energy efficiency of the out-of-condition professor is 9.082 %.

b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Explanation:

a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:

[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)

Where:

[tex]W[/tex] - Useful work, in joules.

[tex]E[/tex] - Food energy, in joules.

If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:

[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]

[tex]\eta = 9.082\,\%[/tex]

The energy efficiency of the out-of-condition professor is 9.082 %.

b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:

[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]

[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]

[tex]E = 7.60\times 10^{5}\,J[/tex]

[tex]E = 181.644\,kcal[/tex]

The food calories needed by the well-conditioned athlete is 181.644 kilocalories.


Related Questions

pha của dao động làm hàm

Answers

Answer:

pha của dao động là hàm bậc nhất của thời gian.

A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.

1. How many revolutions do the kids make before the constant operational speed is reached ?

2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.

3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?

Answers

Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

A uniform magnetic field is directed at an angle of 30o with the plane of a circular coil of radius 4cm and 1000 turns. The magnetic field changes at a rate of 5 T per second. Calculate the following:

a. Area vector
b. Induced emf

Answers

Answer:

(a) The area vector is 0.00503 m² at 30⁰ from the magnetic field

(b) The induced emf is 12.58 V

Explanation:

Given;

angle between the magnetic field and the plane of the circular coil, = 30⁰

number of turns of the coil, N = 1000

radius of the coil, r = 4 cm = 0.04 m

change in the magnetic field with time, dB/dt = 5 T/s

(a) The area vector is calculated as;

A = πr²

A = π x (0.04)²

A = 0.00503 m²

The area vector is 0.00503 m² at 30⁰ from the magnetic field.

(b) The induced emf is calculated as;

[tex]emf = N\frac{\Delta \phi}{\Delta t} \\\\where;\\\\\phi = BAcos \theta\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\where;\\\\\theta \ is \ the \ angle \ between \ a \ perpendicular \ vector \ to \ the \ area\\ and\ the \ magnetic\ field\\\\\theta = 90 - 30 = 60^0\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\emf = (1000) \times (5 )\times (0.00503) \times cos (60)\\\\emf = 12.58 \ V[/tex]

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge

Answers

Answer:

I think it is of science is it true na i knew it bro dont take tension

Work-Energy Theorem & Power
A 0.5 kg mass sitting on smooth ice is accelerated from rest by a force until is
acquires a speed of 8 m/s. The force acts while the mass moves through a
displacement of 2 m.
A. Calculate the kinetic energy of the mass after the force acts.
B. Calculate the work done by the force.
C. Calculate the magnitude of the force that accelerated the mass.

Answers

Answer:

A. 16 J

B. 16 J

C. 8 N

Explanation:

A. Determination of the kinetic energy.

Mass (m) = 0.5 Kg

Velocity (v) =. 8 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 8²

KE = ½ × 0.5 × 64

KE = 0.5 × 32

KE = 16 J

B. Determination of the Workdone by the force.

Kinetic energy (KE) = 16 J

Workdone =.?

Workdone and kinetic energy has the same unit of measurement. Thus,

Workdone = kinetic energy

Workdone = 16 J

C. Determination of the force.

Workdone (Wd) = 16 J

Displacement (s) = 2 m

Force (F) =?

Wd = F × s

16 = F × 2

Divide both side by 2

F = 16 / 2

F = 8 N

A police car travels towards a stationary observer at a speed of 15m/s. the siren on the car emits a sound of frequency 250Hz. Calculate the observer frequency. the speed of sound is 340m/s​

Answers

Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)

= 250 x (340/( 340-15))

= 261.54 Hz

A certain heating element is made out of Nichrome wire and used with the standard voltage source of V=120 V. Immediately after the voltage is turned on, the current running through the element is measured at I1=1.28 A and its temperature at T1=25°C. As the heating element warms up and reaches its steady-state (operating) temperature, the current becomes I2=1.229 A.

Required:
Find this steady-state temperature T2.

Answers

Answer:

T₁ = 232.5 ºC

Explanation:

For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law

           V = i R

            R = V / i

i₀ = 1.28 A

            R₀ = 120 / 1.28

            R₀ = 93.75 ohm

i₁ = 1.229 A

             R₁ = 120 / 1.229

             R₁ = 97.64 or

Resistance in a metal is linear with temperature

            ΔR = α R₀ ΔT

where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹

            ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]

            ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]

            ΔT = 2,075 10² C

            ΔT = T₁-T₀ = 2,075 10²

            T₁ = T₀ + 207.5

             T₁ = 25+ 207.5

             T₁ = 232.5 ºC

If four students separately measure the density of a rock, and they all have very low percent
differences between their measurements, what can you say for certain about the accuracy of their
results?

Answers

Answer:

Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.

Explanation:

Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.

Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.

Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.

Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?

Answers

Answer:

The correct answer is - low pitch

Explanation:

Now for the case it is mentioned that the tube closed on one end frequency is:

f = v/2l

Where,

l = length of the tube

v = velocity of longitudinal wave of gas filled in the tube

if frequency increases then pitch will be increase as well as pitch depends on frequency.

Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.

As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.

A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s.

Answers

This question is incomplete, the complete question is;

A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s. a) θ = 30° V b) θ = 60° V c) θ = 90° V

Answer:

the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

Explanation:

Given the data in the question;

number of turns N = 70

Diameter of coil D = 11 cm

Radius r = D/2 = 11/2 = 5.5 cm = 0.055 m

magnitude of magnetic ΔB = 0.7T

Δt = 0.2 seconds

Now,

a)

For θ = 30°,

Angle of with area of vector θ' = 90° - 30° = 60°

so

emf  e = N( Δ∅ / Δt ) = N( ΔBAcosθ  / Δt )

hence

e =  NAcosθ'(ΔB / Δt )

where A is area ( πr² )

so we substitute

e =  70 × πr² × cos(60°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(60°) × ( 0.7 / 0.2 )

e = 1.16 V

b)

For θ = 60°,

Angle of with area of vector θ' = 90° - 60° = 30°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(30°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.01 V

c)

For θ = 90°,

Angle of with area of vector θ' = 90° - 90° = 0°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(0°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.33 V

Therefore, the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

A uniform circular disk has a radius of 34 cm and a mass of 350 g. Its center is at the origin. Then a circular hole of radius 6.8 cm is cut out of it. The center of the hole is a distance 10.2 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.

Answers

Answer:

u can ask it to the person who give ot to u i dont no

Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field

Answers

Answer:

Option (2) is correct.

The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Explanation:

An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.

So, the correct option is (2).

The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.

The given problem is based on the concept and fundamentals of electromagnetic waves.  The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.

In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.

Learn more about the electromagnetic waves here:

https://brainly.com/question/25559554

The reason why a teacher is more important then a farmer

Answers

Answer:

A teacher is more important than a famer.

Explanation:

A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.

A 900 kg car travelling at 12 m/s due east collides with a 600 kg car travelling at 24 m/s due north. As a result of the collision, the two cars lock together and move in what final direction?

45.0° N of E

53.1° N of E

63.3° N of E

69.5° N of E

Answers

Calculate force of each car:

P1 = 900 kg x 12m/s = 10,800

P2= 600 x 24 = 14,400

Degree of travel = arctan(14,300/10800)

Degree of travel = 53.1 N of E

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop

Answers

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

[tex]V_i=50 \ m/h[/tex]

Now,

[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]

    [tex]=73.333 \ ft/sec[/tex]

Now,

Final velocity,

[tex]V_f=0[/tex]

Initial velocity,

[tex]V_{initial} = 73.333 \ ft/sec[/tex]

hence,

⇒ [tex]V_f^2=V_i^2+2aD[/tex]

By putting the values, we get

      [tex]0=(73.333)^2+2\times( -22) D[/tex]

  [tex]44D=(73.333)^2[/tex]

      [tex]D=\frac{(73.333)^2}{44}[/tex]

          [tex]=122.2211[/tex]

The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns

A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.

Answers

Answer:

C

Explanation:

Similarity

The average 8-18 year old spends how many hours per day average in front of a screen doing little physical activity

Answers

Nearly four hours every day, doing little to no physical activity.

describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please

Answers

Answer:

Force A newton Law first law

F = M.A which Force in 0 N as you Questions Above

Force B

Newton Law 3

Action = -Reaction

Hope you can explain this formula as you want to scribe to explaining

An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m below. Calculate the velocity (in m/s) of the fish just before it hits the water. (Assume that the eagle is flying in the x direction and that the y direction is up.)

Answers

Answer:

Explanation:

The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .

v² = u² + 2 g H .

v² = 0  + 2 x 9.8 x 4.7 m

= 92.12

v = 9.6 m /s

The fish's final velocity will have two components

vertical component = 9.6 m/s downwards

Horizontal component = 2.6 m /s  .

Resultant velocity = √ ( 9.6² + 2.6² )

= √ ( 92.16 + 6.76 )

= 9.9 m /s

Answer:

The speed of fish at the time of hitting the surface is 9.95 m/s.

Explanation:

Horizontal speed, u = 2.6 m/s

height, h = 4.7 m

Let the vertical velocity at the time of hitting to water is v.

Use third equation of motion

[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]

The net velocity with which the fish strikes to the water is

[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]

gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be

Answers

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explain why it takes much more effort to stop a freight train compared with a car?

Answers

Answer:

Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.

Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel

across a 30 V potential difference to a light bulb.

a. Calculate the current delivered through the light bulb in the two cases.

b. Draw the circuit connection that will achieve the brightest light bulb.​

Answers

Explanation:

Given that,

Two resistors 4.5 Ω and 2.3 Ω .

Potential difference = 30 V

When they are in series, the current through each resistor remains the same. First find the equivalent resistance.

R' = 4.5 + 2.3

= 6.8 Ω

Current,

[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]

So, the current through both lightbulb is the same i.e. 4.41 A.

When they are in parallel, the current divides.

Current flowing in 4.5 resistor,

[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]

Current flowing in 2.3 ohm resistor,

[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]

In parallel combination, are brighter than bulbs in series.

An initially motionless test car is accelerated uniformly to 105 km/h in 8.43 s before striking a simulated deer. The car is in contact with the faux fawn for 0.635 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
acceleration before collision:
3.45
m/s2
What is the magnitude of the average acceleration of the car during the collision?
average acceleration during collision:
19.68
m/s2
What is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?

Answers

105 km/h ≈ 29.2 m/s

60.0 km/h ≈ 16.7 m/s

Before the collision the test car has an acceleration a of

a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²

During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is

a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²

i.e. with magnitude about 19.7 m/s².

Overall, the car has an average acceleration of

a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²

The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.

Answers

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         [tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex]            (you has an mistake in the formula)

         [tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]

         [tex]\frac{1}{C_{eq1}}[/tex] = 0.1   10⁶

         [tex]C_{eq1}[/tex] = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          [tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]

          [tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]

          [tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶

          [tex]C_{eq2}[/tex] = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

           

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]

          U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]

          U₄ = 0.5 J

An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16

Answers

Answer:

answer : option (b) 0.016 amp

explanation : resistance of resistor , R = 10 Ω

inductance of inductor , X_LX

L

= 20H

voltage of AC circuit , V = 120volts

frequency, ff =60Hz

so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s

now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=

R

2

2

L

2

V

= 120/√{10² + (120π)² × 20²}

= 120/√{100 + 14400π² × 400}

after solving this we get, i = 0.016 amp

A measurement was made of the magnetic field due to a tornado, and the result was 13.00 nT to the north. The measurement was made at a position 8.90 km west of the tornado. What was the magnitude (in A) and direction of the current in the funnel of the tornado? Assume the vortex was a long, straight wire carrying a current.

Answers

Answer:

4

Explanation:

When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency. The string's tension and mass per unit length remain unchanged.
If the unfingered length of the string is l=65cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12
x1=
x2=
x3=
x4=
x5=
x6=

Answers

Answer:

Explanation:

For frequencies n generated in a string , the expression is as follows

n = 1 /2L√ ( T/m )

n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.

If T and m are constant , then

n x L = constant , hence n is inversely proportional to L or length of string.

Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times

Initial length of string is 65 cm .

x1 = 65 x 1 / 1.75 = 37.14 cm

x2 = 37.14 x 1/ 1.75 = 21.22 cm

x3 = 21.22 x 1 / 1.75 = 12.12 cm

x4= 12.12 x 1 / 1.75 = 6.92 cm

x5 = 6.92 x 1 / 1.75 = 3.95 cm

x6 = 3.95 x 1 / 1.75 = 2.25 cm

hi can anyone pls answer this question. i will mark brainliest​

Answers

Answer:

a.work done by man A is zero

D is best option

Explanation:

D none of them because as we know that to do work some distance should be cover with some load as both picture they are covering some distance with carrying some load so A and B option are absolutely wrong and remaining C they are not doing same amount of work because distance cover by them and load carrying by them is different so how can work be same so D is best option none of A B C is correct answer

A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram

Answers

Answer:

b

Explanation:

The index of refraction for a vacuum is 1.00000. The index of refraction for air is 1.00029. 1) Determine the ratio of time required for light to travel through 1000 m of air to the time required for light to travel through 1000 m of vacuum. (Express your answer to six significant figures.)

Answers

Answer:

 [tex]\frac{t_{air}}{t_{vaccum}}[/tex] = 1.00029

Explanation:

The refractive index is defined

         n = c / v

         v = c / n

the speed of light per se wave is constant, so we can use the relations of uniform motion

         v = x / t

         t = x / v

we substitute

         t = x n / c

let's calculate the time

vacuum

        t₁ = 1000 1/3 10⁸

        t₁ = 3.333333 10⁻⁶ s

air

        t₂ = 1000 1.00029 / 3 10⁸

        t2 = 3.3343 10⁻⁶ s

the relationship between these times is

       t₂ / t₁ = 3.3343 / 3.3333333

       t₂ / t₁ = 1.00029

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