A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position x (t ), of the mass at a function of time, t is

Answer:

the large amplitude wave keeps moving to the right

the small amplitude wave continues to move to the left.

When checking the answers, the correct ones are 1, 2

Explanation:

The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.

The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,

the large amplitude wave keeps moving to the right

the small amplitude wave continues to move to the left.

When checking the answers, the correct ones are 1, 2

Answer:

The thickness is [tex]t = 1.273 *10^{-7} \ m[/tex]  

Explanation:

From the question we are told that

     The  refractive index of the film  is  [tex]n = 1.37[/tex]

      The wavelength is  [tex]\lambda = 696 \ nm = 696 *10^{-9 } \ m[/tex]

Generally the condition for constructive interference in a film is mathematically represented as

        [tex]2 * t = [m + \frac{1}{2} ] \lambda_k[/tex]

Here t is the thickness of the film , m is the order number (0, 1, 2, 3 ... )

[tex]\lambda _k[/tex] is the wavelength of light that is inside the film , this is mathematically evaluated as

       [tex]\lambda _k = \frac{ \lambda }{ n}[/tex]

       [tex]\lambda _k = \frac{ 696 *10^{-9}}{ 1.37}[/tex]

      [tex]\lambda _k = 5.095 *10^{-7 } \ m[/tex]

So  for  m =  0

     [tex]t = [ 0 + \frac{1}{2} ] \lambda _k * \frac{1}{2}[/tex]

substituting values  

  [tex]t = [ 0 + \frac{1}{2} ] (5.095 *10^{-7}) * \frac{1}{2}[/tex]  

  [tex]t = 1.273 *10^{-7} \ m[/tex]  

Answer:

0.6 m/s

Explanation:

2Hz = 2^-1 = 2 /s

30cm = .3m

Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.

(.3m)*(2 /s) = 0.6 m/s

Answer:

3/4 of 12 = 16

3/4 of 24 = 32

Those are the answers based on how your question sounded

Explanation:

Answer:

27

Explanation:

3/4 of (12 and 24)

of means ×

and means +

therefore,3/4× (12+24)

3/4×(36)

3×9

27

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

The correct option is option (A)

the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz

(I) For an organ pipe open at open both ends the frequency of different modes is given by:

F =  nv/2L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = v/2L

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...

F₁ =2v/2L

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

The difference between successive overtones is F₀

(II) For an organ pipe open at one end the frequency of different modes is given by:

F =  nv/4L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

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Answer:The correct answer should be C. is filled with medication used to administer a fixed amount of medication per inhalation through the mouth.

Explanation:

Answer:

c. is filled with medication used to administer a fixed amount of medication per inhalation through the mouth.

Convert to m/s

[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]

We know

[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]

[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]

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.

Answer:

Will be equal to alpha x r; less than UsN

Answer:

Upper plate Q/3

Lower plate 2Q/3

Explanation:

See attached file

Answer:

0.2812

Explanation:

Given that

mass of skater 1, m1 = 45 kg

mass of skater 2, m2 = 60 kg

speed of skater 1, v1 = 0.375 m/s

To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.

Thus, momentum p = mv

Law of conservation of momentum infers that,

m1v1 = m2v2

Now we proceed to substitute our values into the formula.

45 * 0.375 = 60 * v2

v2 = 16.875 / 60

v2 = 0.2812 m/s

Therefore the speed of the second skater has to be 0.2812 m/s

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:

t = T / 2 all energy is stored in the inductor

Explanation:

The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is

          w = √1/LC

          2π / T =√1 / LC

          T = 2π  √LC

The energy is constant and for the initial instant it is completely stored in the capacitor

         Uc = Q₀² / 2C

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor

        U = L I² / 2

in the intermediate instant the energy is stored in the two elements.

Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor

After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,

[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]

here, T is the period of oscillation.

Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,

[tex]U =\dfrac{Q^{2}}{2C}[/tex]

here, C is the capacitance.

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,

[tex]U'=\dfrac{L I^{2}}{2}[/tex]

here, L is the inductance and I is the current.

Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.

Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

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Answer:4 times more energy will be striking the childbearing

Explanation:

Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4

Answer:

 t = 8.98 10⁻⁷ m

Explanation:

This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.

* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180

* The beam when passing to the middle its wavelength changes

           λ = λ₀ / n

if we take this into account, the constructive interference equation for normal incidence is

            2t = (m + ½) λ₀ / n

let's apply this equation to our case

for λ₀ = 479 nm = 479 10⁻⁹ m

             t = (m + ½) 479 10⁻⁹ / 1.33

             (m + ½) = 1.33 t / 479 10⁻⁹

for λ₀ = 798 nm = 798 10⁻⁹ m

             t = (m' + ½) 798 10⁻⁹ /1.33

            (m' + ½) = 1.33 t / 798 10⁻⁹

as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions

               (m + ½) = 1.33 t / 479 10⁻⁹

             ((m-1) + ½) = 1.33 t / 798 10⁻⁹

             (m + ½) = 1.33 t / 798 10⁻⁹ +1

resolve

             1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1

             1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹

             1.33t = (1 .33t + 798 10⁻⁹) 479/798

             1.33t = (1 .33t + 798 10⁻⁹) 0.6

             1.33 t = 0.7983 t + 477.6 10⁻⁹

             t (1.33 - 0.7983) = 477.6 10⁻⁹

             t = 477.6 10⁻⁹ /0.5315

             t = 8.98 10⁻⁷ m

Answer:

The  apparent depth is  [tex]D' = 0.00376 \ m[/tex]

Explanation:

From the question we are told that

     The  depth of the water is  [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]

      The  refractive index of water is  [tex]n = 1.33[/tex]

Generally the apparent depth of the coin is mathematically represented as

          [tex]D' = D * [\frac{ n_a}{n} ][/tex]

Here  [tex]n_a[/tex]  is the refractive index of  air the value is  [tex]n_a = 1[/tex]

So

        [tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]

        [tex]D' = 0.00376 \ m[/tex]

The apparent depth will be 0.00376 m.

The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.

d is the depth of the water =5.0 mm =5.0 ×10⁻³

n is the refractive index of water =1.33

[tex]\rm n_a[/tex] is the refractive index of wire=1

The apparent depth of the coin is given as;

[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]

Hence the apparent depth will be 0.00376 m.

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Answer:

[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]

Given:

Mass of oxygen (m) = 32.0 g = 0.032 kg

Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³

Speed of sound in oxygen (v) = 317 m/s

To Find:

Bulk modulus of oxygen

Explanation:

[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]

[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]

[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]

[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]

[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]

[tex]\sf \implies B = 100489 \times 1.428[/tex]

[tex]\sf \implies B = 143498.292 \: Pa[/tex]

[tex]\sf \implies B \approx 143.5 \: kPa[/tex]

Answer:

The number of windings is 1.

Explanation:

The radius of the solenoid = 8.0 cm = 0.08 m

Length of the solenoid = 45.0 cm = 0.45 m

number of turn = ?

circumference of each winding = 2πr = 2 x 3.142 x 0.08 = 0.503 m

The number of windings = (Length of the solenoid)/(circumference of each winding)

==> 0.45/0.503 = 0.89 ≅ 1

Answer:

The current in the ammeter is zero.

(1) is correct option.

Explanation:

Given that,

The capacitor is charged.

We need find the current after closed switched

We know that,

When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.

Then, the ammeter reads zero.

Hence, The current in the ammeter is zero.

(1) is correct option.

Answer:

A thermos bottle works well because:

A) Its glass walls are thin

Answer:

A thermos bottle works well because:

C

Vacuum reduces heat radiation

Answer:

a)  A = 1.99 μm , b) λ = 0.4134 m , c)  v = 57.2 m / s , d)   s = - 1,946 nm ,

e)      v_max = 1,739 mm / s

Explanation:

A sound wave has the general expression

           s = s₀ sin (kx - wt)

where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is

           s = 1.99 sin (15.2 x - 869 t)

a) the amplitude of the wave is

        A = s₀

        A = 1.99 μm

b) wave spectrum is

      k = 2π /λ

in the equation k = 15.2 m⁻¹

      λ = 2π / k

      λ = 2π / 15.2

     λ = 0.4134 m

c) the speed of the wave is given by the relation

       v = λ f

angular velocity and frequency are related

       w = 2π f

        f = w / 2π

        f = 869 / 2π

        f = 138.3 Hz

        v = 0.4134 138.3

         v = 57.2 m / s

d) To find the instantaneous velocity, we substitute the given distance and time into the equation

       s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

       s = 1.99 sin (0.77368 - 2.55486)

remember that trigonometry functions must be in radians

       s = 1.99 (-0.98895)

       s = - 1,946 nm

The negative sign indicates that it shifts to the left

e) the speed of the oscillating part is

           v = ds / dt)

           v = - s₀(-w) cos (kx -wt)

the maximum speed occurs when the cosines is 1

           v_maximo = s₀w

           v_maximum = 1.99 869

           v_maximo = 1739.31 μm / s

let's reduce to mm / s

          v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

          v_max = 1,739 mm / s

a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s

When A sound wave has the general expression is:

Then, s = s₀ sin (kx - wt)

Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is

s is = 1.99 sin (15.2 x - 869 t)

a) When the amplitude of the wave is

A is = s₀

Thus, A = 1.99 μm

b) When the wave spectrum is

k is = 2π /λ

Now, in the equation k = 15.2 m⁻¹

Then, λ = 2π / k

After that, λ = 2π / 15.2

Thus, λ = 0.4134 m

c) When the speed of the wave is given by the relation is:

Then, v = λ f

Now, the angular velocity and frequency are related is:

w is = 2π f

Then, f = w / 2π

After that, f = 869 / 2π

Now, f = 138.3 Hz

Then, v = 0.4134 138.3

Thus, v = 57.2 m / s

d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation

Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

After that, s = 1.99 sin (0.77368 - 2.55486)

Then remember that trigonometry functions must be in radians

After that, s = 1.99 (-0.98895)

Thus, s = - 1,946 nm

When The negative sign indicates that it shifts to the left

e) When the speed of the oscillating part is

Then, v = ds / dt)

Now, v = - s₀(-w) cos (kx -wt)

When the maximum speed occurs when the cosines is 1

Then, v_maximo = s₀w

After that, v_maximum = 1.99 869

v_maximo = 1739.31 μm / s

Now, let's reduce to mm / s

Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

Therefore, v_max = 1,739 mm / s

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Answer:

The wave is travelling in the ±z-axis direction.

Explanation:

An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the  electric and magnetic field.

In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.

Answer:

Explanation:

speed of light ≈ 300 000 000 m/s

a) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is nearest Earth?

(230e9 - 150e9) / 3e8 = 277 s or 4 minutes 27 seconds.

b) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is farthest from Earth?

(230e9 + 150e9) / 3e8 = 1267 s or 21 minutes 6 seconds.

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

R(parallel) = product/ sum

50×30/50+30

1500/80

18,75 ohms

Answer: 18.75

above is right but you need to put a dot after the number 18

Answer:

A) it transforms a small force acting over a large distance into a large force acting over a small distance.

Explanation:

The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.

Pressure transmitted P = F/A

where F is the force applied

and A is the area over which the force is applied.

This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.

F/A = f/a

where the left side of the equation is for the output, and the right side is for the input.

The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that

volume V = (area A) x (distance d)

this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.

From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.

Answer:

The focal length of the lens is 2.5 cm

Explanation:

Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;

[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]

Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:

[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]

then, solving for the focal distance knowing that the object distance is 1.5 cm:

[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/tex]

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

a)

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

τ = 4900 N.m

b)

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

τ = 4900 N.m

c)

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

τ = 4900 N.m

Hence, the correct answer will be:

e. The torque is the same for all cases.

Calculate the horizontal distance travelled by a ball throw with a velcoity 20 sqrt 2 ms^(-1) without hitting the ceiling of an anditorium of heith 20 m. Use g= 10 ms^(-2). =(20√2)2sin2×45∘10=9800×1)10=80m .

HOPE SO IT HELPS YOU

Answer:

The same amount of current flows through the battery and light bulb

Explanation:

Because for a single loop, the current is the same at every point in the loop. Thus, the amount of current that flows through the lightbulb is the same as the amount that flows through the battery

Answer:

The same amount of current flows through the battery and light bulb

Explanation: