You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .

Requried:
What is the magnitude of the charge (in nC) on each bead?

Answer:

T_object = 380.35 K

Explanation:

From Stefan–Boltzmann law, the power output is given by the formula:

P = σAT⁴

where;

σ is Stefan-Boltzmann constant

A is area of the radiating surface.

T is temperature of the body

Now, we are told that the power the object emitted is 3 times the power absorbed from the room.

Thus, we have;

P_e = 3P_a

Where P_e is power emitted and P_a is power absorbed.

So, we have;

σA(T_object)⁴ = 3σA (T_room)⁴

σA will cancel out to give;

(T_object)⁴ = 3(T_room)⁴

We are given T_room = 289 K

Thus;

(T_object)⁴ = 3 × 289⁴

(T_object) = ∜(3 × 289⁴)

T_object = 380.35 K

Answer:

Explanation:

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

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

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

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

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

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

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

Given parameters

Np = 500-turns

Ns = 2000-turns

Is = 3.0Amp

Required

Current in the primary coil (Ip)

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

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

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

Hence the current in the primary coil is 12Amp

Answer:

The actual height is 3.308 m.

Explanation:

The person is swimming below the water surface at distance = 0.80 m  

The height of the diving board appears at a distance or height = 5.20 m

Now we have to find the actual distance of the diving board from the water surface.

We know the refractive index of water is 1.33.

Therefore, the actual height = (Distance that appears – distance below the water surface) / Refractive index.

The actual height = ( 5.20 - 0.80 ) / 1.33 = 3.308 m

Explanation:

It is given that,

Distance covered by the cart to the right is 2 m

Distance covered by the cart to the left is 1 m

We need to find the total distance rolled by the cart. Total distance is equal to the sum of the distances covered by an object. It does depend on the direction.

Total distance = 2 m + 1 m

D = 3 m

The cart rolled to a total distance of 3 m.

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

We have that for the Question, it can be said that the average induced emf in the coil is

E=0.028565V

From the question we are told

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?

Generally the equation for the Average emf induced   is mathematically given as

[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]

Hence

[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]

Therefore

[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]

E=0.028565V

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Answer:

F = 2.7×10¯⁶ N.

Explanation:

From the question given:

F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²

Thus we can obtain the value value of F by carrying the operation as follow:

F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²

F = 2.7648×10¯⁷ / 0.1024

F = 2.7×10¯⁶ N.

Therefore, the value of F is 2.7×10¯⁶ N.

....................

Answer:

b. Third Law entropy  

Explanation:

Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature"  tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.

In the question above, the correct answer is option b.

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m

Answer:

far away

Explanation:

There are different types of eye defect ranging from short sightedness, longsighted, astigmatism, presbyopia etc.

If someone is only able to see close ranged object clearly but not far distant object, then such person is suffering from short sightedness or myopia. This occurs when the light rays entering the eye does not converge on the retina. Instead of converging on the retina, the light ray is formed on a point in front of the retina. This causes the distance from the eye's lens to the retina shorter compared to that of a normal eye. This eye defect is usually corrected using concave lens in order to diverge the rays thereby allowing it to focus on the retina.

Hence, if the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are far away (at a far distant).

Explanation:

According to Ohms Law :

V = I * R

(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms

Also,

[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]

(B)

[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]

Drift speed is missing. It is given as;

1.7 × 10^(-5) m/s

A) R = 0.0293 ohms

B) ρ = 1.589 × 10^(-8)

C) n = 8.8 × 10^(28) electrons

This is about finding, resistance and resistivity.

Length; L = 5.8 m

Diameter; d = 2mm = 0.002 m

Radius; r = d/2 = 0.001 m

Voltage; V = 22 mv = 0.022 V

Current; I = 750 mA = 0.75 A

Area; A = πr² = 0.001²π

Drift speed; v_d = 1.7 × 10^(-5) m/s

R = V/I

R = 0.022/0.75

R = 0.0293 ohms

ρ = RA/L

ρ = (0.0293 × 0.001²π)/5.8

ρ = 1.589 × 10^(-8)

J = n•e•v_d

Where;

J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²

e is charge on an electron = 1.6 × 10^(-19) C

v_d = 1.7 × 10^(-5) m/s

n is number of free electrons per unit volume

Thus;

238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))

238732.44 = (2.72 × 10^(-24))n

n = 238732.44/(2.72 × 10^(-24))

n = 8.8 × 10^(28)

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Answer:

a) the distance between the interference fringes is reduced by half

b) the distance between stripes is doubled

Explanation:

Interference experiments constructive interference is described by the expression

          d sin θ = m λ

let's use trigonometry to find the distance between the interference fringes

              tan θ=  y / L

dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles

            tan θ = sin θ / cos θ = sin θ

we substitute

             sin θ = y / L

            d y / L = m  λ

           y = m λ / d L

a) it asks us when the screen doubles its distance

           L ’= 2 L

subtitute in the equation

           y ’= m λ / (d 2L)

           y ’=( m λ / d L) /2

           y ’= y / 2

the distance between the interference fringes is reduced by half

b) the density of the network doubles

      if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half

            d ’= d / 2

we substitute

          y ’= m λ (L d / 2)

          y ’= m λ / (L d) 2

          y ’= y 2

the distance between stripes is doubled

Answer:

the net work is zero

Explanation:

Work is defined by the expression

        W = F. ds

Bold type indicates vectors

In this problem, the friction force does not decrease, therefore it will be zero.

Consequently for work on a closed path it is zero.

The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero

Answer:

The value is  [tex]\mu = 0.76[/tex]

Explanation:

From the question we are told that

    The  acceleration is [tex]a = 7.4 \ m /s^2[/tex]

Generally the force by which the truck bed (truck) is moving with is mathematically represented as

          [tex]F = ma[/tex]

Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet  is equal the force by which the the truck bed is moving with that is  

        [tex]F_f = F[/tex]

Here  [tex]F_f[/tex] is the frictional force which is mathematically represented as

         [tex]F_f = \mu * m * g[/tex]

substituting into above equation

         [tex]\mu * m * g = ma[/tex]

=>        [tex]\mu = \frac{a}{g}[/tex]

substituting values

           [tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]

           [tex]\mu = 0.76[/tex]

Explanation:

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

And, from equation ( 2 ), we get

[tex]\theta=\frac{x}{d}[/tex]

Thus,

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

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

[tex]=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}[/tex]

[tex]=0.0515 \mathrm{m}[/tex]

Thus, the diameter of the telescope's mirror that would allow us to see details as small as is

Answer:

λmax = 372 nm

Explanation:

First we find the energy of photon:

E = hc/λ

where,

E = Energy of Photon = ?

λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m

c = speed of light = 3 x 10⁸ m/s

h = Planks Constant = 6.626 x 10⁻³⁴ J.s

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)

E = 8.5 x 10⁻¹⁹ J

Now, from Einstein's Photoelectric Equation:

E = Work Function + Kinetic Energy

8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)

Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J

Work Function = 5.332 x 10⁻¹⁹ J

Since, work function is the minimum amount of energy required to emit electron. Therefore:

Work Function = hc/λmax

λmax = hc/Work Function

where,

λmax = maximum wavelength of light that will produce photoelectrons = ?

Therefore,

λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)

λmax = 3.72 x 10⁻⁷ m

λmax = 372 nm

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

Answer

The effect is that it Decreases the field current IF and increases slope K1

Explanation:

gravitional force attracts objects with mass toward each other.

The correct answer is B. Calcite

Explanation:

Mohs hardness scale indicates the hardness of minerals using a scale from 1 to 10 as well as defining the objects or tools that can be used to scratch the minerals. These two features of minerals are shown in the table of the image. About this, it is shown gypsum and talc can be scratched by just a fingernail, considering minerals with a hardness of 2.5 or below can be scratched by a fingernail. In the case of calcite that has a hardness of 3, this cannot be scratched by a fingernail, but it can be scratched by a penny, which works for minerals with a hardness of 3.5 or below. Thus, the correct answer is Calcite.

Answer:

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Explanation:

Given:

Frequency = 91.5 MHz

Power = 11.5 Kw = 11,500 J/s

Find:

Number of photons emit per second

Computation:

Total energy with frequency (E) = hf

Total energy with frequency (E) = 6.626×10⁻³⁴  × 91.5×10⁶

Total energy with frequency (E) = 6.06×10⁻²⁶ J

Number of photons emit per second = 11,500 / 6.06×10⁻²⁶

Number of photons emit per second = 1897.689 × 10²⁶

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Answer:

a

   [tex]\lambda = 1.667 nm[/tex]

b

     [tex]\theta = 0.8681^o[/tex]

Explanation:

From the question we are told that

   The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]

    The  is distance of the screen from the slit is  [tex]D = 2.60 \ m[/tex]

    The distance between the central bright fringe and either of the adjacent bright   [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]

Generally  the condition for constructive interference is  

      [tex]d sin \tha(\theta ) = n \lambda[/tex]

From the question we are told that small-angle approximation is valid here.

So    [tex]sin (\theta ) = \theta[/tex]

=>        [tex]d \theta = n \lambda[/tex]

=>        [tex]\theta = \frac{n * \lambda }{d }[/tex]

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         [tex]y = D * sin (\theta )[/tex]

From the question we are told that small-angle approximation is valid here.

So

       [tex]y = D * \theta[/tex]

=>   [tex]\theta = \frac{ y}{D}[/tex]

So

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

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

substituting values

       [tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]

        [tex]\lambda = 1.667 *10^{-6}[/tex]

        [tex]\lambda = 1.667 nm[/tex]

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     [tex]dsin (\theta ) = n \lambda[/tex]

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

    [tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]

    [tex]\theta = 0.8681^o[/tex]

Answer:

The radius is  [tex]r = 3.1905 \ m[/tex]

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  [tex]d = 2.70 \ m[/tex]

        The refractive index of the liquid is  [tex]n_i = 1.31[/tex]

Now the critical value is mathematically represented as

         [tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]

substituting values

         [tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]

         [tex]\theta = 49.76^o[/tex]

Using SOHCAHTOA rule we have that

         [tex]tan \theta = \frac{ r}{d}[/tex]

=>     [tex]r = d * tan \theta[/tex]

substituting values  

        [tex]r = 2.7 * tan (49.76)[/tex]

        [tex]r = 3.1905 \ m[/tex]

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= - 4 m/s

Answer:

Unbalenced

Explanation:

when balenced forces are applied to an object there is no motion. When you apply unbalenced force the object you are applying the force to will move in the opposite direction of the force.

Answer:

im pretty sure it unbalenced

Explanation:

i just am

(a) The car is undergoing an acceleration of

[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]

so that in 9.02 s, it will have covered a distance of

[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]

The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:

[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]

(b) The wheels have average angular velocity

[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]

where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.

The wheels start at rest, so

[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]

Answer:

8 mm

Explanation:

From the information given:

The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.

when the distance is less than the radius ; we have:

[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]

when the distance is greater than the radius; we have:

[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]

Equating both equations together ; we have :

[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]

[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]

[tex]R= \dfrac{d^2}{r}[/tex]

where; d = radius of the wire and r = distance;

[tex]R =\dfrac{4^2}{2}[/tex]

[tex]R =\dfrac{16}{2}[/tex]

R = 8 mm

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

The apparent depth of the fish is 60 cm.

To calculate the apparent depth of the fish, we use the formula below.

Formula:

Where:

Make D' The subject of the equation.

From the question,

Given:

Substitute these values into equation 2

Hence, the apparent depth of the fish is 60 cm

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Explanation:

The frequency is given to be f = 8 Hz.

Period is the inverse of frequency.

T = 1/f = 0.125 s

Velocity is wavelength times frequency.

v = λf = (0.40 m) (8 Hz) = 3.2 m/s

The wave travels 3.2 meters every second.

Answer:

The power of the mirror in diopters is 58.8 D

Explanation:

Given;

radius of curvature of the spoon, R = 3.4 cm = 0.034 m

The focal length of a mirror is given by;

[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]

The focal length of the mirror is 0.017 m

(b) The power of the mirror is given by;

[tex]P = \frac{1}{f}[/tex]

where;

P is the power of the mirror

f is the focal length

[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]

Thus, the power of the mirror in diopters is 58.8 D