You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 110 kg, perfectly absorbing block.

Required:
a. What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
b. Does this seem like a promising method for launching satellites?

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

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:

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:

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:

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:

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:

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:

  P₂ = 2 P₁

we conclude that in the second time the power used is double that in the first rise

Explanation:

In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.

Now we can analyze the required power,

         P = W / t

From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,

therefore the first time the power is

           P₁ = E / 10

           P₁ = 0.1 E

for the second time the power is

          P₂ = E / 5

          P₂ = 0.2 E

we see that the power in the second case is

         P₂ = 2 P₁

Therefore, we conclude that in the second time the power used is double that in the first rise.

Explanation:

Elongation of the wire is:

ΔL = F L₀ / (E A)

where F is the force,

L₀ is the initial length,

E is Young's modulus,

and A is the cross sectional area.

ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle.  There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

ΔL = 0.00167 m

ΔL = 1.67 mm

Answer:

B) approaches infinity

Explanation:

The capacitive reactance of an AC capacitor is given by;

[tex]X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}[/tex]

Where;

C is the capacitance

f is the frequency of the ac voltage

[tex]X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity[/tex]

Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.

The correct option is (B) approaches infinity

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]

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

Learn more about mechanical energy from here

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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: 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:

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:

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:

4.003" (inches )

Explanation:

The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .

Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is

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:

Its not A..

Explanation:

I chose A - was incorrect

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 magnitude of the acceleration is 5.11 m/s²

Explanation:

Given;

Tension on the rope, T = 164 N

mass of the bucket, m = 11 kg

Weight of the rope is given by;

W = mg = 11 x 9.8 = 107.8 N

According to Newton's second law of motion, the tension on the rope is given by;

T = W - ma

ma = W - T

ma = 107.8N - 164N

ma = -56.2 N

a = -56.2 / m

a = -56.2 / 11

a = -5.11 m/s²

The magnitude of the acceleration is 5.11 m/s²

Answer:

Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.

Difference:

Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.

Answer:

(a) When the resistance R is doubled, I = 1 A

(b) When the peak emf εo is doubled, I = 4 A

(c)  When the frequency ω is doubled, I = 2 A

Explanation:

Given;

peak current through the resistor, I = 2.0 A

According to ohms law the peak current through the circuit is given by;

[tex]I = \frac{V}{R}[/tex]

(a) When the resistance R is doubled;

[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]

(b)When the peak emf εo is doubled

[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]

(c) When the frequency ω is doubled

Peak current through resistor is independent of frequency

I₂ = 2.0 A

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 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:

Pressure, P = 3724 Pa

Explanation:

Given that,

Depth of water, [tex]h_w=20\ cm =0.2\ m[/tex]

Depth of oil, [tex]h_o=40-20=20\ cm=0.2\ m[/tex]

The density of water, [tex]d_w=1000\ kg/m^3[/tex]

The densinty of oil, [tex]d_o=900\ kg/m^3[/tex]

We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :

[tex]P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa[/tex]

So, the gauge pressure at the bottom of the cylinder is 3724 Pa.

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

Explanation:

Given that,

Mass of a box is 2.6 kg

(1) We need to find the work done on the box by this force as it is pushed up the 5.00-m ramp to a height of 3.00 m.

It means that the position of the object is 3 m i.e. h = 3 m

Work done = Fd

= mgh

So,

[tex]W=2.6\times 9.8\times 3\\\\W=76.44\ J[/tex]

(2) Now the gravitational potential energy that the box undergoes as it rises to its final height is equal to the work done by the box. So,

Change in potential energy = 76.44 J

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:

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]