A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the lens.
Part A. How far must the lens be from the photocells?
s = cm
Part B. Is the image on the photocells erect or inverted? Real or virtual?
a. The image is erect and real.
b. The image is inverted and real.
c. The image is erect and virtual.
d. The image is inverted and virtual.
Part C. How tall is the image on the photocells?
|h?| = cm
Part D. A SLR digital camera often has pixels measuring 8.00?m

Answers

Answer 1

Answer:

a. 7.62cm

b. Real and inverted

c. 2.76 cm

d. 3450

Explanation:

We proceed as follows;

a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;

1/f = 1/p + 1/q

making q the subject of the formula;

q = pf/p-f

From the question;

p = 4.70m

f = 7.5cm = 0.075m

Substituting these, we have ;

q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm

b. The image is real and inverted since the image distance is positive

c. We want to calculate how tall the image is

Mathematically;

h1 = (q/p)h0

h1 = (7.62/4.70)* 1.7

h1 = 2.76 cm

d. We want to calculate the number of pixels that fit into this image

Mathematically:

n = h1/8 micro meter

n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450


Related Questions

"Light traveling in a medium with a refractive index 1.11 is incident on a plate of another medium with index of refraction 1.66. At what angle of incidence is the reflected light fully polarized?"

Answers

Answer:

56°

Explanation:

Brewsters angle can be simply derived from

n1sin theta1= n2sintheta2= n2costheta1

because the reflected light will be 100% polarized if it is reflected at an angle 90o to the refracted light. Hence, Brewsters angle is

Tan theta= n2/n1

1.66/1.11= 1.495

Theta = 56°

Explanation:

Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s

Answers

Answer:

The time difference is 2.97 sec.

Explanation:

Given that,

Tension = 29 N

Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]

Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]

Length of first section = 2.5 m

Length of second section = 2.8 m

We need to total distance of first pulse

Using formula for distance

[tex]d=2.5+2.5[/tex]

[tex]d_{1}=5.0\ m[/tex]

We need to total distance of second pulse

Using formula for distance

[tex]d=2.8+2.8[/tex]

[tex]d_{2}=5.6\ m[/tex]

We need to calculate the speed of pulse in the first string

Using formula of speed

[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]

Put the value into the formula

[tex]v_{1}=\sqrt{\dfrac{29}{19}}[/tex]

[tex]v_{1}=1.24\ m/s[/tex]

We need to calculate the speed of pulse in the second string

Using formula of speed

[tex]v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}[/tex]

Put the value into the formula

[tex]v_{2}=\sqrt{\dfrac{29}{45}}[/tex]

[tex]v_{2}=0.80\ m/s[/tex]

We need to calculate the time for first pulse

Using formula of time

[tex]t_{1}=\dfrac{d_{1}}{v_{1}}[/tex]

Put the value into the formula

[tex]t_{1}=\dfrac{5.0}{1.24}[/tex]

[tex]t_{1}=4.03\ sec[/tex]

We need to calculate the time for second pulse

Using formula of time

[tex]t_{2}=\dfrac{d_{1}}{v_{1}}[/tex]

Put the value into the formula

[tex]t_{2}=\dfrac{5.6}{0.80}[/tex]

[tex]t_{2}=7\ sec[/tex]

We need to calculate the time difference

Using formula of time difference

[tex]\Delta t=t_{2}-t_{1}[/tex]

Put the value into the formula

[tex]\Delta t=7-4.03[/tex]

[tex]\Delta t=2.97\ sec[/tex]

Hence, The time difference is 2.97 sec.

Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?

Answers

Answer:

The wavelength is  [tex]\lambda = 1805 nm[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]

     The  distance of the screen is  D  =  3.0  m

     The  fringe width is  [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]

     

Generally the fringe width for a bright fringe  is mathematically represented as

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

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

=>     [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]

=>     [tex]d = 0.000373 \ m[/tex]

Generally the fringe width for a dark fringe  is mathematically represented as

      [tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

Here  m = 0  for  first order dark fringe

   So  

         [tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

looking at which we see that   [tex]y_d = y[/tex]

         [tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]

=>    [tex]\lambda = 1805 *10^{-9} \ m[/tex]

=>    [tex]\lambda = 1805 nm[/tex]

In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 540 nmnm . Part A What is the work function of this material

Answers

Answer:

Φ = 36.84 × 10^(-20) J

Explanation:

In the photoelectric effect, the energy of the incoming photon is usually used in part to extract the photoelectron from the material (work function) and then the rest is converted into kinetic energy of the photoelectron which is given by the formula;

K_max = hf - Φ

where;

hf represents the energy of the incoming photon

h is the Planck's constant

f is the light frequency

Φ is the work function of the material

K_max is the maximum kinetic energy of the photoelectrons.

From the question, we are told that no current flows unless the wavelength is less than 540 nm. This means that when the wavelength has this value, the maximum kinetic energy of the photoelectrons is zero i.e K_max = 0. Thus the energy of the incoming photons is just enough to extract the photoelectrons from the material.

Thus,

hf - Φ = 0

hf = Φ - - - (1)

We are given the wavelength as;

λ = 540 nm = 540 × 10^(-9) m

Now, let's find the frequency of the light by using the relationship between frequency and wavelength. The equation is;

f = c/λ

Where c is speed of light = 3 × 10^(8) m/s

f = (3 × 10^(8))/(540 × 10^(-9))

f = 5.56 × 10^(14) Hz

Thus, from equation 1,we can now find the work function;

Φ = hf

h is Planck's constant and has a value of 6.626 × 10^(-34) J.s

Thus;

Φ = 6.626 × 10^(-34) × 5.56 × 10^(14)

Φ = 36.84 × 10^(-20) J

An elderly sailor is shipwrecked on a desert island but manages to save his eyeglasses. The lens for one eye has a power of 1.28 diopters, and the other lens has a power of 8.50 diopters. What is the magnifying power of the telescope he can construct with these lenses

Answers

10.88 is the answer to the question

Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.

Answers

Answer:

81 N

7.1 m/s²

Explanation:

Draw a free body diagram of the sphere.  There are two forces:

Weight force mg pulling straight down,

and tension force T pulling up along the rope.

Sum of forces in the centripetal direction:

∑F = ma

T − mg sin 45° = m v² / r

T = m (g sin 45° + v² / r)

T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)

T = 81 N

Sum of forces in the tangential direction:

mg cos 45° = ma

a = g cos 45°

a = (10 m/s²) cos 45°

a = 7.1 m/s²

What is the impedance of an AC series circuit that is constructed of a 10.0-W resistor along with 12.0 W inductive reactance and 7.0 W capacitive reactance

Answers

Answer:

11.2 Ω

Explanation:

The impedance of a circuit is given by;

Z= √R^2 +(XL-XC)^2

Since

Resistance R= 10 Ω

Inductive reactance XL= 12 Ω

Capacitive reactance XC= 7 Ω

Z= √10^2 + (12-7)^2

Z= √100 + 25

Z= √125

Z= 11.2 Ω

During the first part of this lab, we want to determine how the object distance is related to what two quantities

Answers

Answer and Explanation:

The computation of the object distance related to two quantities is shown below:

It could find out by using the lens formula which is shown below:

[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]

where,

v = image distance

u = object distance

f = focal length

It could be found by applying the above formula i.e considering the image distance, object distance and the focal length

In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 W/m2 at the floor of the facility.
A) Find the average radiation pressure (in pascals) on a totally absorbing section of the floor.B) Find the average radiation pressure (in atmospheres) on a totally absorbing section of the floor.C) Find the average radiation pressure (in pascals) on a totally reflecting section of the floor.D) Find the average radiation pressure (in atmospheres) on a totally reflecting section of the floor.

Answers

Answer:

a) 8.33 x 10^-6 Pa

b) 8.23 x 10^-11 atm

c) 1.67 x 10^-5 Pa

d) 1.65 x 10^-10 atm

Explanation:

Intensity of the light [tex]I[/tex] = 2500 W/m^2

speed of light [tex]c[/tex] = 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

[tex]P_{abs}[/tex] = [tex]I/c[/tex] = 2500/(3 x 10^8) = 8.33 x 10^-6 Pa

b) 1 atm = 101325 Pa

[tex]P_{abs}[/tex] = (8.33 x 10^-6)/101325 = 8.23 x 10^-11 atm

c) for a totally reflecting surface

[tex]P_{ref}[/tex] = [tex]2I/c[/tex] = twice the value for totally absorbing

[tex]P_{ref}[/tex]  = 2 x 8.33 x 10^-6 = 1.67 x 10^-5 Pa

d)  1 atm = 101325 Pa

[tex]P_{ref}[/tex] = 2 x 8.23 x 10^-11  = 1.65 x 10^-10 atm

If a ray of light traveling in the liquid has an angle of incidence at the interface of 33.0 ∘, what angle does the refracted ray in the air make with the normal?

Answers

Answer:

29°

Explanation:

because the refracted ray angle is small than angle of incidence

A fish is 11.9 cm from the front surface of a fish bowl of radius 33 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl

Answers

Answer:

The fish would appear 42.7 cm on the left side from the front of the bowl.

Explanation:

The fish (object) distance = 11.9 cm, radius of curvature of the bowl = 33 cm. The distance of image of the fish (image distance) can be determined by applying the mirror formula;

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

where f is the focal length of the reflecting surface, u is the object distance and v is the image distance.

But, f = [tex]\frac{radius of curvature}{2}[/tex]

         = [tex]\frac{33}{2}[/tex]

       f = 16.5 cm

Substitute f = 16.5 = [tex]\frac{165}{10}[/tex], and u = 11.9 = [tex]\frac{119}{10}[/tex] in equation 1;

[tex]\frac{10}{165}[/tex] = [tex]\frac{10}{119}[/tex] + [tex]\frac{1}{v}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{10}{165}[/tex] - [tex]\frac{10}{119}[/tex]

  = [tex]\frac{1190 - 1650}{19635}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{-460}{19635}[/tex]

⇒ v  = [tex]\frac{19635}{-460}[/tex]

       = -42.6848

    v = 42.7 cm

The fish would appear 42.7 cm on the left side from the front of the bowl.

Heat and thermodynamics Numerical ​

Answers

Answer:

K = 227.04 W/m.°C

Explanation:

First we need to find the heat required to melt the ice:

q = m H

where,

q = heat required = ?

m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg

H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg

Therefore,

q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)

q = 2839 J

Now, we find the heat transfer rate through rod:

Q = q/t

where,

t = time = (10 min)(60 s/1 min) = 600 s

Q = Heat Transfer Rate = ?

Therefore,

Q = 2839 J/600 s

Q = 4.73 W

From Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

K = Thermal Conductivity = ?

A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²

L = Length of rod = 60 cm = 0.6 m

ΔT = Difference in temperature = 100°C - 0°C = 100°C

Therefore,

4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m

K = (4.73 W)/(0.0208 m.°C)

K = 227.04 W/m.°C

How does the direction of current flow in the coil affect the orientation of the magnetic field produced by the electromagnet

Answers

Answer:

The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.

Determine the final angular velocity of a particle that rotates 4500 ° in 3 seconds and an angular acceleration of 8 Rad / s ^ 2

Answers

Answer:

the final angular velocity of the particle is approximately 38.18  Rad/s

Explanation:

To start with, let's make sure that units of angle measure are the same, converting everything into radians:

[tex]4500^o\, \frac{\pi}{180^o}= 25\,\pi[/tex]

And now we can use the kinematic formulas for rotational motion:

[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2[/tex]

Therefore we can find the initial angular velocity [tex]\omega_0[/tex]  of the particle:

[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2\\25\,\pi=\omega_0\,(3)+\frac{1}{2} (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=\frac{25\,\pi-36}{3} \\\omega_0\approx 14.18\,\,\,rad/s[/tex]

and now we can estimate the final angular velocity using the kinematic equation for angular velocity;

[tex]\omega=\omega_0\,+\alpha\,t\\\omega=14.18+8\,(3)\\\omega=38.18\,\,\,rad/s[/tex]

An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated

Answers

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s

Answers

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

Calculation of Wavelength

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

Finf more informmation about Wavelength here:

https://brainly.com/question/6352445

A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).

Answers

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

[tex]B = \frac{N\mu_o I}{2R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻ T.

An airplane propeller is rotating at 2200 rpm . You may want to review (Pages 255 - 259) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Rotation of a compact disc.
A. How many seconds does it take for the propeller to turn through 49.0?
t = 4.41x10^-3 S
B. Compute the propeller's angular velocity in rad/s
w = 194 rad/s

Answers

Answer:

a) 3.7 x 10^-3 s

b) 230.41 rad/s

Explanation:

The angular speed N = 2200 rpm (revolution per minute)

==> 2200/60 revolutions per sec = 36.67 rps

The total angle turned in one second = 36.67 x 360° = 13201.2°

if it takes 1 sec to revolve 13201.2°

then it will take t sec to rotate 49.0°

time t = 49/13201.2 = 3.7 x 10^-3 s

conversion to rad/s = 2πN/60 = (2 x 3.142 x 2200)/60 = 230.41 rad/s

PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.

Answers

Answer:

1.The temperature of each sample will increase by the same amount

Explanation:

This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.

This is shown by the calculation below

Q = mcΔT

ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.

Since Q, m and c are the same for both substances, thus ΔT will be the same.

So, the temperature of each sample will increase by the same amount

Convert 7,348 grams to kilograms

Answers

0.00735 kilograms is your answer. I’m sorry If I didn’t explain it right :(,

Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh's criterion

Answers

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

     

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              [tex]D = \frac{1.22 \lambda }{d }[/tex]

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.65 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.

Answers

Answer:

2.2 ft

Explanation:

0.65 lb / 1 ft = 1.4 lb / x

x ≈ 2.2 ft

A body is thrown vertically upwards with a speed of 95m / s and after 7s it reaches its maximum height. How fast does it reach its maximum height? What was the maximum height reached?

Answers

Explanation:

u = 95 m/sec ( Initial speed)

t = 7 sec ( Time of ascent)

According to Equations of Motion :

[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]

Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m

Answer:

332.5 m

Explanation:

At the maximum height, the velocity is 0.

Given:

v₀ = 95 m/s

v = 0 m/s

t = 7 s

Find: Δy

Δy = ½ (v + v₀) t

Δy = ½ (0 m/s + 95 m/s) (7 s)

Δy = 332.5 m

You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *

Answers

Answer:

 cost = $ 243.00

Explanation:

This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m

                 #_tables = 3 m (1 / 0.20 m)

                #_tables = 15 tables

Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?

              Cost = 15 (16.20 / 1)

              cost = $ 243.00

A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire

Answers

Answer:

The direction of the force is towards the East.

Explanation:

Using the right hand rule, the force on the current carrying conductor is east.

In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.

In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.

. Two waves that have the same wavelengths and amplitudes are traveling in opposite directions on a string. If each wave has a speed of 10 m/s and there are moments when the string is not moving, what is the wavelength of the waves if the time between each moment that the string is flat is 0.5 s?

Answers

Answer:

10m

Explanation:

Since Given frequency f= 1/t

and velocity ν=10 m/s

We know ν=λf

λ= ν/f

​ = 10/1/0.5

=5m

Since both the waves are similar but moves in opposite direction its total wavelength of the wave will be 10 m

The isotope (_90^234)Th has a half-life of 24days and decays to (_91^234)Pa. How long does it take for 90% of a sample of (_90^234)Th to decay to (_91^234)Pa?

Answers

Answer:

79.7 days

Explanation:

Half-life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is the amount of time,

and T is the half life.

If 90% decays, then 10% is left.

A = A₀ (½)^(t / T)

0.1 A₀ = A₀ (½)^(t / 24)

0.1 = ½^(t / 24)

ln(0.1) = (t / 24) ln(0.5)

t ≈ 79.7 days

A 23 cm tall object is placed in front of a concave mirror with a radius of 37 cm. The distance of the object to the mirror is 86 cm. Calculate the focal length of the mirror.

Answers

Answer:

18.5 cm

Explanation:

From;

1/u + 1/v = 1/f

Where;

u= object distance = 86cm

image height = 23 cm

Radius of curvature = 37 cm

The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.

Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm

Therefore, the focal length of the mirror is 18.5 cm

One solenoid is centered inside another. The outer one has a length of 54.0 cm and contains 6750 coils, while the coaxial inner solenoid is 4.00 cm long and 0.170 cm in diameter and contains 21.0 coils. The current in the outer solenoid is changing at 35.0 A/s .What is the mutual inductance of the solenoids?Find the emf induced in the inner solenoid.

Answers

Answer:

 M₁₂ = 1.01 10⁻⁴ H ,   Fem = 3.54 10⁻³ V

Explanation:

The mutual inductance between two systems is

        M₁₂ = N₂ Ф₁₂ / I₁

where N₂ is the number of turns of the inner solenoid N₂ = 21.0, i₁ the current that flows through the outer solenoid I₁ = 35.0 A / s and fi is the flux of the field of coil1 that passes through coil 2

         

the magnetic field of the coil1 is

   B = μ₀ n I₁ = μ₀ N₁/l   I₁

the flow is

             Φ = B A₂

the area of ​​the second coil is

             A₂ = π d₂ / 4

             Φ = μ₀ N₁ I₁ / L  π d² / 4

we substitute in the first expression

            M₁₂ = N₂ μ₀ N₁ / L    π d² / 4

            M₁₂ = μ₀ N₁ N₂ π d² / 4L

           d = 0.170 cm = 0.00170 m

            L = 4.00 cm = 0.00400 m

let's calculate

            M₁₂ = 4π 10⁻⁷ 6750  21 π 0.0017²/ (4 0.004)

             M₁₂ = π² 0.40966 10⁻⁷ / 0.004

             M₁₂ = 1.01 10⁻⁴ H

The electromotive force is

              Fem = - M dI₁ / dt

              Fem = - 1.01 10⁻⁴ 35.0

              Fem = 3.54 10⁻³ V

A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination.​ Question 96 options:

Answers

Answer:

"Endoscope" is the correct answer.

Explanation:

A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.
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