A 19.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.8-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.22 s .

Required:
What is the average induced emf in the loop?

Answer:

λ = 4.1638 10⁻⁷ m

Explanation:

The photoelectric effect was explained by Einstein assuming that the radiation acts like particles and the equation that describes the process is

           K = h f -Ф

where K is the kinetic energy of the emitted electrons, hf  the energy of the photons according to Planck's equation and Ф the work function of the material

In this case they give us the kinetic energy of the electrons

         K = 0.7 eV

The sodium work function is tabulated Ф = 2.28 eV

Let's find the frequency of the photons

            f = (K + Ф) / h

Planck's constant is

          h = 6.626 10⁻³⁴ J s (1 eV / 1.6 10⁻¹⁹ J) = 4.136 10⁻¹⁵ eV s

            f = (0.7 + 2.28) / 4.136 10⁻¹⁵

            f = 7.2050 10¹⁴ Hz

let's find the wavelength using the relationship between speed and frequency and wavelength

            c = λ f

            λ = c / f

            λ = 3 10⁸ / 7.205 10¹⁴

            λ = 4.1638 10⁻⁷ m

Her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.

As given in the problem A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg in total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%

The potential energy is getting converted into kinetic energy with an efficiency of 52 %

1/2mv² = 0.52 (mgh)

v²= 1.04gh

v = √(1.04gh)

v= √(1.04×9.81×10)

v = 10.1 m/s

Thus, her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be  10.1 m/s

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Answer:

p ≈ f_ objective     Therefore for the object to be in focus it must be close to the focal length

Explanation:

A microscope is an optical instrument that uses two lenses, or a long focal length objective lens that forms a real image of the object and an eyepiece that forms a virtual image of the object. Therefore for the object to be in focus it must be close to the focal length

           p ≈ f_ objective

p distance objetive

Answer:

The index of refraction of quartz for violet light is 1.47.

Explanation:

It is given that, a narrow beam of white light is incident on a sheet of quartz.

The beam disperses in the quartz, with red light at an angle, [tex]\theta_r=26.3^{\circ}[/tex] wrt to the normal and violet light traveling at an angle of [tex]\theta_v=25.7^{\circ}[/tex]

The index of refraction of quartz for red light is 1.45.

We need to find the index of refraction of quartz for violet light.

Using Snell's law of red light as follows :

[tex]\mu_a\sin\theta_i=\mu_r\sin\theta_r[/tex]

Here,

[tex]\mu_a[/tex] is the refractive index of air

[tex]\theta_i[/tex] is the angle of incidence

We can find the value of angle of incidence as follows :

[tex]\sin\theta_i=\dfrac{\mu_r \sin\theta_r}{\mu_a}\\\\\sin\theta_i=\dfrac{1.45\times \sin(26.3)}{1}\\\\\theta_i=\sin^{-1}(0.642)\\\\\theta_i=39.79^{\circ}[/tex]

Now again using Snell's law for violet light as follows :

[tex]\mu_a\sin\theta_i=\mu_v\sin\theta_v\\\\\mu_v=\dfrac{\mu_a\sin\theta_i}{\sin\theta_v}\\\\\mu_v=\dfrac{1\times \sin(39.79)}{\sin(25.7)}\\\\\mu_v=1.47[/tex]

So, the index of refraction of quartz for violet light is 1.47.

Answer: Wavelength of 424 nm  can produce photoelectrons from silver

Explanation:

[tex]\phi=h\times \nu_o=\frac{hc}{\lambda}[/tex]

[tex]\phi[/tex] = work function = energy of photon

h = Planck's constant =  [tex]6.63\times 10^{-34}Js[/tex]

[tex]\nu_0[/tex] = frequency of the metal

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

[tex]\lambda[/tex] =longest  wavelength of the radiation

Now put all the given values in the above formula, we get the wavelength of the photons.

[tex]\lambda=\frac{hc}{\phi}[/tex]

[tex]\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^8m/s}{2.93\times 1.6\times 10^{-19}J}[/tex]      ( as 1ev=[tex]1.6\times 10^{-19}J[/tex] )

[tex]\lambda=4.24\times 10^{-7}m[/tex]

[tex]1nm=10^{-9}m[/tex]

[tex]\lambda=424nm[/tex]

Here, wavelength of 424 nm  can produce photoelectrons from silver

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation: The Graphite was the material in the passage that had reflected most of the light.

Answer:

It does not depend on direction of plane and the left windtips more potential

Explanation:

Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction

Answer:

a

  [tex]n = 23[/tex]

b

  [tex]v = 87377.95 \ m/s[/tex]

Explanation:

From the question we are told that

   The diameter is [tex]d = 61\ nm = 61 *10^{-9} \ m[/tex]

Generally the radius electron orbit  is mathematically represented as

      [tex]r = \frac{61 *10^{-9}}{2}[/tex]

=>   [tex]r = 3.05*10^{-8} \ m[/tex]

This radius can also be represented mathematically  as

      [tex]r = n^2 * a_o[/tex]

Here n is the quantum number and [tex]a_o[/tex] is  the Bohr radius with a value

    [tex]a_o = 0.0529 *10^{-9} \ m[/tex]

So

   [tex]n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }[/tex]

=>   [tex]n = 23[/tex]

Generally the angular momentum of the electron is mathematically represented as

          [tex]L = m * v * r = \frac{n * h }{2 \pi}[/tex]

Here  h is the Planck constant and the value is  [tex]h = 6.626*10^{-34} J \cdot s[/tex]

          m is the mass of the electron with values [tex]m = 9.1*10^{-31} \ kg[/tex]

         So

               [tex]v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }[/tex]

                [tex]v = 87377.95 \ m/s[/tex]

Answer: She adds different amounts of the chemical to the material and then puts them in the water

Explanation:

Answer: One group of fabric is treated with the chemical, and the other group is not. Then each group is exposed to water.

Explanation:

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

Answer:

0.006<357<700.003<6010<9256.0<9520.00

Answer:

90 hp

Explanation:

Power = work / time

P = ½ (1500 kg) (25 m/s)² / 7.0 s

P = 67,000 W

P = 90 hp

Answer:

17,300 m

Explanation:

Using kinematic equations, first find the time it takes to land.

Δy = v₀ t + ½ at²

0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²

t = 0 s or 68.5 s

The horizontal distance it moves in that time is:

Δx = v₀ t + ½ at²

Δx = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²

Δx = 17,300 m

Alternatively, you can use the range equation:

R = v₀² sin(2θ) / g

R = (420 m/s)² sin(2 × 53.0°) / (9.8 m/s²)

R = 17,300 m

The distance a cannonball will land if it is fired on flat ground at 420 m/s at a 53.0° angle is 17,300 meters.

The complete movement of an object, regardless of direction, is referred to as distance. The amount of ground a thing travels from its starting point to its destination is also referred to as distance.

Given:

A cannonball is fired on flat ground at 420 m/s at a 53.0° angle,

Calculate the time to land on the ground as shown below,

[tex]\Delta y = v_o t +1/2 at^2[/tex]

0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²

t = 0 s or 68.5 s

Calculate the distance as shown below,

[tex]\Delta x = v_o t +1/2 at^2[/tex]

Δ x = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²

Δ x = 17,300 m

Thus, the total distance covered by the cannonball fired with a speed of 420 m/s is 17300 meters.

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Answer:

The viscosity of the fluid is 1.16 N-s/m²

Explanation:

Given that,

Area = 0.6 m²

Angle = 30°

Velocity = 0.36 m/s

Thickness = 1.8 mm

Weight = 280 N

We need to calculate the viscosity of the fluid

Using balance equation

[tex]w\sin\theta=\dfrac{\mu\times v}{t}\times A[/tex]

Put the value in the equation

[tex]280\sin30=\dfrac{\mu\times0.36}{1.8\times10^{-3}}\times(0.6)[/tex]

[tex]140=\mu\times120[/tex]

[tex]\mu=1.16\ N-s/m^2[/tex]

Hence. The viscosity of the fluid is 1.16 N-s/m².

Answer:

The time taken is  [tex]t = 0.356 \ s[/tex]

Explanation:

From the question we are told that

  The length of steel the wire is  [tex]l_1 = 31.0 \ m[/tex]

   The  length of the  copper wire is  [tex]l_2 = 17.0 \ m[/tex]

    The  diameter of the wire is  [tex]d = 1.00 \ m = 1.0 *10^{-3} \ m[/tex]

     The  tension is  [tex]T = 122 \ N[/tex]

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              [tex]t = t_s + t_c[/tex]

Where  [tex]t_s[/tex] is the time taken to transverse the steel wire which is mathematically represented as

         [tex]t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ][/tex]

here  [tex]\rho_s[/tex] is the density of steel with a value  [tex]\rho_s = 8920 \ kg/m^3[/tex]

   So

      [tex]t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]

      [tex]t_s = 0.235 \ s[/tex]

 And

        [tex]t_c[/tex] is the time taken to transverse the copper wire which is mathematically represented as

      [tex]t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ][/tex]

here  [tex]\rho_c[/tex] is the density of steel with a value  [tex]\rho_s = 7860 \ kg/m^3[/tex]

 So

      [tex]t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]

      [tex]t_c =0.121[/tex]

So  

   [tex]t = t_c + t_s[/tex]

    [tex]t = 0.121 + 0.235[/tex]

    [tex]t = 0.356 \ s[/tex]

Answer:

1.29*10^6 N/C

1135.6 V

9.18 cm

Explanation:

Given that

radius of the metal, r = 19 cm

charge of the metal, q = 2.4*10^-8 C

coulomb's constant, k = 8.99*10^9

to find the electric field, we use the formula E = kq/r², where

E = electric field

k = coulomb constant

q = charge on the metal and

r = radius of the metal

E = (8.99*10^9 * 2.4*10^-8) / 0.19²

E = 215.76 / 0.0361

E = 1.29*10^6 N/C

to find the electric potential, we use this relation

V = kq/r

V = (8.99*10^9 * 2.4*10^-8) / 0.19

V = 215.76 / 0.19

V = 1135.6 V

V = kq/r,

r = kq/V

r = (8.99*10^9 * 2.4*10^-8)/ (1135.6 - 370

r = 215.76 / 765.6

r = 0.281 = 28.1 cm

distance from the sphere

28.18 - 19 = 9.18 cm

Answer:

After 80 years there will be 3 g of element X remaining

Explanation:

Given;

the half life of element X = 20 year

initial mass of element X = 48 g

a) How much is there after 80 years

0 year --------------------------> 48 g

20 years -----------------------> (48g / 2) = 24 g

40 years ------------------------> 12 g

60 years ------------------------> 6 g

80 years --------------------------> 3 g

Therefore, after 80 years there will be 3 g of element X remaining.

Answer:

250 m

Explanation:

1. First, we have to calculate the acceleration:

[tex]\boxed{\mathsf{v=v_o+a\cdot t}}[/tex]

where:

v = present velocity

vo = initial velocity

a = acceleration

t = time

2. Let us use given information and substitute in the expression above. Hence:

[tex]\mathsf{50=0+a\cdot10}\\\\\mathsf{50=10\cdot a}\\\\\mathsf{a=\dfrac{50}{10}}\\\\\therefore \boxed{\mathsf{a=5\,m/s^2}}}[/tex]

3. Now we can calculate traveled distance with Torricelli's equation:

[tex]\boxed{\mathsf{v^2=v_o^2+2\cdot a \cdot d}}[/tex]

where:

v = present velocity

vo = initial velocity

a = accelerarion

d = distance

4. So, we get:

[tex]\mathsf{50^2=0^2+2\cdot 5 \cdot d}\\\\\mathsf{2500=10\cdot d}\\\\\mathsf{d=\dfrac{2500}{10}}\\\\\therefore \boxed{\mathsf{d=250\,m}}[/tex]

Conclusion: during 10 seconds the truck has traveled a distance of 250 m.

Have a nice day! : )

If a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds. During the 10 seconds, the truck traveled a distance of 250 meters.

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

By using the first equation of motion,

v = u + at

50 = 0 + 10 a

a = 50/10

a= 5 meters/second²

As given in the problem a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds.

The total distance traveled by truck,

S = ut + 0.5at²

S = 0 + 0.5 ×5 ×10²

S = 250 meters

Thus, the total distance traveled by truck would be 250 meters.

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Answer:

The  value is  [tex]v = -0.04 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass  of the block is  [tex]m = 2.0 \ kg[/tex]

   The  force constant  of the spring is  [tex]k = 590 \ N/m[/tex]

   The amplitude  is  [tex]A = + 0.080[/tex]

   The  time consider is  [tex]t = 0.10 \ s[/tex]

Generally the angular velocity of this  block is mathematically represented as

      [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]w = \sqrt{\frac{590}{2} }[/tex]

=>   [tex]w = 17.18 \ rad/s[/tex]

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         [tex]v = -A w sin (w* t )[/tex]

=>       [tex]v = -0.080 * 17.18 sin (17.18* 0.10 )[/tex]

=>       [tex]v = -0.04 \ m/s[/tex]

The velocity of the block at the given time is -0.04 m/s.

The angular speed of the block is determined by using the following wave equation;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{590}{2} } \\\\\omega = 17.176 \ rad/s[/tex]

The velocity of the block at the given time is calculated as follows;'

[tex]v = - A sin(\omega t)\\\\v =- 0.08 \times sin(17.176 \times 0.1)\\\\v = -0.04 \ m/s[/tex]

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Answer:

u = 88.54 m/s

Explanation:

Given that,

A discus thrower achieves a high throw of 100 m.

Angle of projection is 30°

We need to find the initial speed of the discuss.​ It is a cse of projectile motion. The maximum height reached by the discus is given by :

[tex]H=\dfrac{u^2\sin^2\theta}{2g}[/tex]

u is the initial speed of the discus

So,

[tex]u^2=\dfrac{2\times 9.8\times 100}{\sin^2(30)}\\\\u=\sqrt{7840}\\\\u=88.54\ m/s[/tex]

So, the initial speed of the discus is 88.54 m/s.

Answer:

The cart's acceleration is [tex]\approx 3.71\,\,\frac{m}{s^2}[/tex]

Explanation:

Let's start by finding the net force acting on the cart, and then find its acceleration using Newton's 2nd Law.

Net force = 95.4 N -36.0 N = 59.4 N

Now, since we know the cart's mass, we can use Newton's 2nd Law to find the cart's acceleration:

[tex]F=m\,*\,a\\a=\frac{F}{m} \\a=\frac{59.4}{16} \,\,\frac{m}{s^2} \\a\approx 3.71\,\,\frac{m}{s^2}[/tex]

Answer:

3.71 m/s²

Explanation:

Answer:

1.41eV

Explanation:

Kinetic Energy of photoelectron(K. Emax)

E = Wo + K.Emax

E = hc/ λ

h = planck's constant = 6.63 * 10^-34

c = speed of light = 3×10^8 m/s

λ = 400nm

Work function (Wo) = 1.70eV

1 eV = 1×10^-19

E = [(6.63×10^-34) * (3×10^8)] / 400×10^-7

E = (19.89 × 10^(-26))/400×10^-7

E = 0.049725×10^-19

K.Emax = E - Wo

K.Emax = (0.049725×10^-19) - (1.7×10^-38)

0.049725×10^-19 interms of eV = (0.049725/1.6)×10^-19 =

K.Emax = 3.11eV - 1.70eV

K.Emax = 1.41eV

Answer: His average speed in mph over the last 400 m is 7.7 m/s.

Explanation:

Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.

We know that , speed = [tex]\dfrac{distance}{time}[/tex]

Here , distance = 400m and time = 51.9 s

Then, speed =  [tex]\dfrac{400}{51.9}\approx7.7\ m/s[/tex]

Hence, his average speed in mph over the last 400 m is 7.7 m/s.

Answer: 17.2 MPH

Explanation:

Average speed = distance/time

400m÷ 51.9s= 7.7 m/s

Now convert m/s →MPH

m→km→mi and s→min→hr

[tex]\frac{7.7m}{s} x[/tex] [tex]\frac{1 km}{1000 m} x\frac{0.6214 mi}{1 km} x\frac{60 s}{1 min} x\frac{60 mins}{1 hr} = 17.2 mph[/tex]

Answer:

The mass of the chunk of gold is 241.25 g

Explanation:

Since density is defined as mass divided by volume, we can solve for the mass (m) via its equation:

[tex]density=\frac{mass}{volume} \\19.3=\frac{m}{12.5} \\m=19.3\,(12.5)\, grams\\m = 241.25 \,\,g[/tex]

Answer:

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Explanation:

Given;

mass of first marble, m₁ = 50g = 0.05 kg

initial velocity of the first marble, u₁ = 2 m/s

mass of second marble, m₂ = 20 g = 0.02 kg

initial velocity of the second marble, u₂ = 0

final velocity of the first marble, v₁ = 1 m/s

Let the final velocity of the second marble, = v₂

Determine the final velocity of the second marble by applying principle of momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂

0.1 = 0.05 + 0.02v₂

0.02v₂ = 0.1 - 0.05

0.02v₂ = 0.05

v₂ = 0.05 / 0.02

v₂ = 2.5 m/s

During inelastic collision both objects will move at the same velocity after collision.

During elastic collision both objects will move at different velocities after collision.

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Answer:

The value of the inductance is 4.75 H

Explanation:

Given;

initial current through the inductor, I₁ = 2.4 A

final current through the inductor, I₂ = 0.3 A

duration of change of current, dt = 1.75 s

voltage change of the inductor, V = 5.7 Volts

The voltage change of the inductor is given by;

[tex]V_L = -L\frac{di}{dt}\\\\ V_L = -L(\frac{I_2-I_1}{dt} )\\\\V_L = L(\frac{I_1-I_2}{dt} )\\\\[/tex]

Where;

L is the inductance of the coil;

[tex]5.7 = L(\frac{2.4-0.3}{1.75} )\\\\5.7 = 1.2 L\\\\L = \frac{5.7}{1.2}\\\\ L = 4.75 \ H[/tex]

Therefore, the value of the inductance is 4.75 H

Answer:

The inductance is  [tex]L = 0.1097 \ H[/tex]

Explanation:

From the question we are told that

  The  length is  [tex]l = 0.65 \ m[/tex]

   The  diameter is  [tex]d = 3.2 cm = 0.032 \ m[/tex]

    The  number of loops is  [tex]N = 8400[/tex]

Generally the radius is evaluated as

       [tex]r = \frac{ 0.032 }{2} = 0.016 \ m[/tex]

The  inductance is mathematically represented as

      [tex]L = \frac{ \mu_o * N^2 * A }{ l }[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

A is the cross-sectional area which is mathematically evaluated as

             [tex]A = \pi r^2[/tex]

=>        [tex]A = 3.142 * (0.016)^2[/tex]

=>        [tex]A = 0.000804 \ m^2[/tex]

 =>  [tex]L = \frac{ 4\pi * 10^{-7} * 8400^2 * 0.000804 }{ 0.65 }[/tex]

=>    [tex]L = 0.1097 \ H[/tex]

Answer:

the pressure fluctuation is LONGITUDINAL

Explanation:

Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.

The expression for the wave is

        ΔP = Δo sin (kx - wt)

Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL

Answer:

The number of turns of the solenoid is 1217 turns

Explanation:

Given;

current in the solenoid, I = 2 A

length of the solenoid, L = 34 cm = 0.34 m

magnitude of the magnetic field, B = 9 mT = 0.009 T

Number of turns of the solenoid = N

The magnitude of magnetic field at the center of the solenoid is given by;

B = μnI

Where;

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

I is the current in the solenoid

n is the number of turns per length

n = B/μI

n = (0.009) / (4π x 10⁻⁷)(2)

n = 3580.52 turns/m

N = nL

N =(3580.52 turns/m) x (0.34 m)

N = 1217 turns

Therefore, the number of turns of the solenoid is 1217 turns