Find the rms current delivered by the power supply when the frequency is very large. Answer in units of A.

Answer:

Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Explanation:

The length of the telescope is

         L = f_ocular + f_objetive

the magnification of the telescope is

         m = - f_objective / f_ocular

These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

[tex]E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A[/tex]

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

[tex]E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV[/tex]

Therefore, the average emf induced in the coil is 175 mV

Answer:

b) the heavier one will have twice the kinetic energy of the lighter one.

Explanation:

The kinetic energy of object with mass, m

K.E₁ = ¹/₂mv²

where;

m is mass of the object

v is the velocity of the object

Since, the two objects are falling under same acceleration due to gravity, their velocity will be increasing at the same rate

The kinetic energy of object with mass, 2m

K.E₂ = ¹/₂(2m)v²

K.E₂ = 2(¹/₂mv²)

BUT K.E₁ = ¹/₂mv²

K.E₂ = 2(K.E₁)

Therefore, the heavier one will have twice the kinetic energy of the lighter one.

b) the heavier one will have twice the kinetic energy of the lighter one.

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

[tex]Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h[/tex]

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

[tex]h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}[/tex]

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

[tex]Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000[/tex]

Pb' = 122 KPa

(A)  The depth of the fluid is 2.14 m.

(B)  The new pressure at the bottom of container is 121972 Pa.

Given data:

The cross-sectional area of the container is, [tex]A =66.2 \;\rm cm^{2}=66.2 \times 10^{-4} \;\rm m^{2}[/tex].

The density of fluid is, [tex]\rho = 856 \;\rm kg/m^{3}[/tex].

The container pressure at bottom is, [tex]P=119 \;\rm kPa=119 \times 10^{3} \;\rm Pa[/tex].

The atmospheric pressure is, [tex]P_{at}=101 \;\rm kPa=101 \times 10^{3}\;\rm Pa[/tex].

(A)

The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,

[tex]P_{net} = P-P_{at}\\\\\rho \times g \times h= P-P_{at}[/tex]

Here, h is the depth of fluid.

Solving as,

[tex]856\times 9.8 \times h= (119-101) \times 10^{3}\\\\h=\dfrac{ (119-101) \times 10^{3}}{856\times 9.8}\\\\h= 2.14 \;\rm m[/tex]

Thus, the depth of the fluid is 2.14 m.

(B)

For an additional volume of [tex]2.35 \times 10^{-3} \;\rm m^{3}[/tex] to the liquid, the new depth is,

[tex]V=A \times h'\\\\h'=\dfrac{2.35 \times 10^{-3}}{66.2 \times 10^{-4}}\\\\h'=0.36 \;\rm m[/tex]

Now, calculate the new pressure at the bottom of the container as,

[tex]P'-P_{at}= \rho \times g \times (h+h')\\\\\P'-(101 \times 10^{3})= 856 \times 9.8 \times (2.14+0.36)\\\\P'=121972 \;\rm Pa[/tex]

Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.

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Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turns of the coil, 200 turn

cross sectional area of the coil, A = 300 cm² = 0.03 m²

magnitude of the magnetic field, B = 30 mT = 0.03 T

Maximum value of the induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

Where;

ω is angular frequency = 2πf

E = NBA(2πf)

f = E / 2πNBA

f = (8) / (2π x 200 x 0.03 x 0.03)

f = 7.07 Hz

Therefore, the frequency of the coil is 7.07 Hz

Answer:

At the end of the handle farthest from the head of the hammer.

Explanation:

The force of the hammer is greatest the longer the radius is on a which would be the length of the handle. Simple mechanical advantage.

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

[tex] B = \frac{\mu_{0}NI}{L} [/tex]

[tex] \frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m [/tex]

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

[tex] N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns [/tex]

Since each turn has length 2πr of wire, the total length is:

[tex] L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m [/tex]

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

Answer:

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;

[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = [tex]\frac{1}{(2/15)}[/tex]

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength [tex]\lambda[/tex] = 2m

Transverse speed [tex]v = f \lambda[/tex]

[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]

Hence, the transverse speed at that point is  15m/s

Answer:

221 lines per millimetre

Explanation:

We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.

Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,

tanθ = 6.09 cm/42.0 cm = 0.145

From trig ratios, cot²θ + 1 = cosec²θ

cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969

sinθ = 1/cosecθ = 1/6.969 = 0.1435

Also, sinθ = mλ/d at the first-order maximum, m = 1. So

sinθ = (1)λ/d = λ/d

Equating both expressions we have  

0.1435 = λ/d

d = λ/0.1435

Now, λ = 650 nm = 650 × 10⁻⁹ m

d = 650 × 10⁻⁹ m/0.1435

d = 4529.62 × 10⁻⁹ m per line

d = 4.52962 × 10⁻⁶ m per line

d = 0.00452962 × 10⁻³ m per line

d = 0.00452962 mm per line

Since d = width of grating/number of lines of grating

Then number of lines per millimetre = 1/grating spacing

= 1/0.00452962

= 220.77 lines per millimetre

≅ 221 lines per millimetre since we can only have a whole number of lines.

Answer:

See attached file

Explanation:

Answer:

The amplitude of the resultant wave is 12.93 cm.

Explanation:

The amplitude of resultant of two waves, y₁ and y₂, is given as;

Y = y₁ + y₂

Let y₁ = A sin(kx - ωt)

Since the wave is out phase by φ, y₂ is given as;

y₂ = A sin(kx - ωt + φ)

Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )

Given;

phase difference, φ = 45°

Amplitude, A = 7.00 cm

Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )

Y = 12.93 cm

Therefore, the amplitude of the resultant wave is 12.93 cm.

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, [tex]d_o[/tex] = 15.0 cm

image distance, [tex]d_i[/tex] = 5.0 cm

height of the object, [tex]h_o[/tex] = 9.0 cm

height of the image, [tex]h_i[/tex] = ?

Apply lens equation;

[tex]\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm[/tex]

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

Answer:

From fastest speed to slowest speed, the electromagnetic waves are ranked as(up to down):

d. X-ray

c. Green light

a. Yellow light

f. Infrared wave

b. FM radio wave

e. AM radio wave

Explanation:

Electromagnetic waves are waves produced as a result of vibrations between an electric field and a magnetic field. The waves have three properties and these properties are frequency, speed and wavelength, which are related by the relationship below

V = Fλ

where:\

V = speed (velocity)

F = frequency

λ = wavelength.

From the relationship above, it is seen that the speed of a wave is directly proportional to its frequency. The higher the frequency, the higher the speed. Therefore, from the list given, the waves with  the highest to lowest frequencies/ from left to right are:

X-ray (3×10¹⁹ Hz to 3×10¹⁶Hz), Green light (5.66×10¹⁴Hz), Yellow light (5.17×10¹⁴Hz), Infrared wave (3×10¹¹Hz), FM radio wave (10.8×10⁸Hz to 8.8×10⁷Hz), AM radio wave (1.72 × 10⁶Hz to 5.5×10⁵Hz).

This corresponds to the speed from highest to lowest from left to right.

Answer:

d.

Explanation:

Since, the capacitance( decreases )

therefore voltage between the plates(increases ).

Hence, option d is correct.

C =εA/d.

d is doubled, therefore  C decrease ( inverse relation).

D) The capacitance decreases and the voltage between the plates increases.

A battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the plates is doubled without loss of charge. Accordingly, the capacitance decreases and the voltage between the plates increases.

The capacitance - (decreases)

The voltage between the plates- (increases ).

Thus, the correct answer is D.

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Answer:

a

   [tex]F = 67867.2 \ N[/tex]

b

  [tex]\Delta L = 2.6 \ mm[/tex]

Explanation:

From the question we are told that

      The Young modulus is  [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]

      The stress is  [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]

      The  diameter is  [tex]d = 2.40 \ cm = 0.024 \ m[/tex]

The radius is mathematically represented as

       [tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]

The cross-sectional area is  mathematically evaluated as

        [tex]A = \pi r^2[/tex]

         [tex]A = 3.142 * (0.012)^2[/tex]

        [tex]A = 0.000452\ m^2[/tex]

Generally the stress is mathematically represented as

        [tex]\sigma = \frac{F}{A}[/tex]

=>     [tex]F = \sigma * A[/tex]

=>    [tex]F = 1.50 *10^{8} * 0.000452[/tex]

=>    [tex]F = 67867.2 \ N[/tex]

Considering part b

      The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]

Generally Young modulus is mathematically represented as

           [tex]E = \frac{ \sigma}{ strain }[/tex]

Here strain is mathematically represented as

         [tex]strain = \frac{ \Delta L }{L}[/tex]

So    

       [tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]

        [tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]

=>     [tex]\Delta L = \frac{\sigma * L }{E}[/tex]

substituting values

       [tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]

       [tex]\Delta L = 0.0026[/tex]

Converting to mm

      [tex]\Delta L = 0.0026 *1000[/tex]

      [tex]\Delta L = 2.6 \ mm[/tex]

Answer:

Acceleration of the body is:

[tex]a=0.27\,\,m/s^2[/tex]

Explanation:

Use Newton's second Law to solve for the acceleration:

[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]

Answer:

The velocity is  [tex]v_h = 19.2 \ m/s[/tex]

Explanation:

From the question we are told that

   The speed of the roller coaster at ground level  is [tex]v = 26 \ m/s[/tex]

Generally we can define the roller coaster speed at ground level using the an equation of motion as

     [tex]v^2 = u^2 + 2 g s[/tex]

 u is  zero given that the roller coaster started from rest

      So

            [tex]26^2 = 0 + 2 * g * s[/tex]

So  

           [tex]s = \frac{26^2}{ 2 * g }[/tex]

=>       [tex]s = 37.6 \ m[/tex]

Now the displacement half way is mathematically represented as

    [tex]s_{h} = \frac{37.6}{2}[/tex]

     [tex]s_{h} = 18.8 \ m[/tex]

So

      [tex]v_h ^2 = u^2 + 2 * g * s_h[/tex]

Where  [tex]v_h[/tex] is the velocity at the half way point

=>  [tex]v_h = \sqrt{ 0 + 2 * 9.8 * 18.8 }[/tex]

=>   [tex]v_h = 19.2 \ m/s[/tex]

Answer:

[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]

Explanation:

Two forces are acting on the object.

Subtracting 2 N from both forces.

2 N → Object ← 5 N

- 2 N                 - 2N

0 N → Object ← 3 N

The force 3 N is pushing the object to the left side.

The mass of the object is 10 kg.

Applying formula for acceleration (Newton’s Second Law of Motion).

a = F/m

a = 3/10

a = 0.3

Answer:

Thomas is correct that the zero displacements

Lilian is right that the distance is greater than zero.

Explanation:

In this problem we have to be clear about the difference between displacement and distance.

The displacement is a vector, that is, it has a modulation and direction, in this case we can draw a vector for the outward trip and another vector for the return trip, both will have the same magnitude, but their directions are opposite, so the resulting vector is zero.

The distance is a scalar and its value coincides with the modulus of the distance vector, in our case the distance is d for the outward journey and d for the return journey, so the total distance is 2d, which is different from zero.

The two students have some reason, but neither complete,

The displacement is zero because it is a vector and

the distance is different from zero (2d) because it is a scalar

Thomas is correct that the zero displacements

Lilian is right that the distance is greater than zero.

Therefore I agree with both, because each one has a 50% of the reason

Answer:

v = 345.6m/s

Explanation:

v = 384 x 0.9 = 345.6

v = 345.6m/s

Answer:

The width is [tex]w_c = 0.00422 \ m[/tex]

Explanation:

From the question we are told that

   The  wavelength is  [tex]\lambda = 6.33*10^{-7} \ m[/tex]

    The  width of the slit is  [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]

    The distance of the screen is  [tex]D = 1.0 \ m[/tex]

Generally the central maximum is mathematically represented as

      [tex]w_c = 2 * y[/tex]

Here  y is the width of the first order maxima which is mathematically represented as

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

substituting values

      [tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]

       [tex]y = 0.00211 \ m[/tex]

So  

    [tex]w_c = 2 *0.00211[/tex]

     [tex]w_c = 0.00422 \ m[/tex]

Answer:

The time taken is  [tex]\Delta t = 1.5 \ s[/tex]

Explanation:

From the question we are told that

   The mass of the ball is  [tex]m = 1.25 \ kg[/tex]

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  [tex]\theta = 1 rev = 2 \pi \ radian[/tex]

Generally the displacement for one  complete revolution is mathematically represented as

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

Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]

     [tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]

     [tex]\alpha = 0.9698 \ s[/tex]

Now the displacement for two  complete revolution is

         [tex]\theta_2 = 2 * 2\pi[/tex]

         [tex]\theta_2 = 4\pi[/tex]

Generally the displacement for two complete revolution is mathematically represented as  

     [tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]

=>   [tex]t^2 = 25.9187[/tex]

=>   [tex]t= 5.1 \ s[/tex]

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     [tex]\Delta t = t_2 - t[/tex]

substituting values

      [tex]\Delta t = 5.1 - 3.60[/tex]

     [tex]\Delta t = 1.5 \ s[/tex]

The time for the ball to complete the next revolution is 1.5 s.

The given parameters;

The constant angular acceleration of the ball is calculated as follows;

[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]

The time to complete the next revolution is calculated as follows;

[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]

[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]

Thus, the time for the ball to complete the next revolution is 1.5 s.

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Answer:

No

Explanation:

Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.

:-)

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

Answer:

Explanation:

We shall apply formula of Doppler's effect

Here source is fixed and observer is approaching the source

f = f₀ x [(V + v ) / V ]

f₀ is original and f is apparent frequency , V is velocity of sound and v is velocity of motorcyclist .

f = 1000 x [(331 + 27.27 ) / 331 ]

= 1082 .4 Hz

This is the frequency heard by motorcyclist .

When police car is approaching him when he is stopped

f = f₀ x [V /(V - v ) ]

v is velocity of police car .

= 1000  x 331 / (331 - 27.27)

= 1090 Hz  

Answer:

The current is  [tex]I = 8.9 *10^{-5} \ A[/tex]

Explanation:

From the question we are told that

     The  radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]

      The current density is  [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]

      The distance we are considering is  [tex]r = 0.5 R = 0.001585[/tex]

Generally current density is mathematically represented as

          [tex]J = \frac{I}{A }[/tex]

Where A is the cross-sectional area represented as

         [tex]A = \pi r^2[/tex]

=>      [tex]J = \frac{I}{\pi r^2 }[/tex]

=>    [tex]I = J * (\pi r^2 )[/tex]

Now the change in current per unit length is mathematically evaluated as

        [tex]dI = 2 J * \pi r dr[/tex]

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         [tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]

         [tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]

substituting values

        [tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]

        [tex]I = 8.9 *10^{-5} \ A[/tex]

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

A) Applying the energy equation

B) The energy output by the athlete is ; 800 Joules

C) The metabolic power is : 2000 w

D) When he performs the task in 1.2 s

Given data :

Weight of barbell = 400 N

Height = 2.0 m

Time = 1.6 secs

efficiency of the human body = 25%

Speed = constant

A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its  potential energy increases.  ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement

B)  Determine the energy output of the athlete

weight of barbell * Height  = 400 * 2 = 800 J

C) Determine the metabolic power

Metabolic power = energy output / Time

where ; energy output = 4 * 800 = 3200

∴ Metabolic power = 3200 / 1.6

                                = 2000 w

D) When performs same task at 1.2 s

The metabolic energy he expends is  the same  and His metabolic power is  more

Hence we can conclude that the answers to your questions are as listed above

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Answer;

26.45cm

See attached file for explanation