The work function of a certain metal is φ = 3.55 eV. Determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (Planck's constant is: h = 4.1357×10-15 eVs.)

Answer:

The time taken is  [tex]t = 2.739 *10^{8} \ s[/tex]

Explanation:

From the question we are told that

    The speed of the spacecraft is [tex]v = 0.99c[/tex]

    where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]

    =>   [tex]v = 0.99 * 3.0 *10^{8 } = 2.97*10^{8}\ m/s[/tex]

    The distance of Sirius is [tex]d = 8.6 \ light-years = 8.6 * 9.461*10^{15}= 8.135*10^{16} \ m[/tex]

Generally the time taken is mathematically represented as

       [tex]t = \frac{d}{v}[/tex]

substituting values

      [tex]t = \frac{8.136 *10^{16}}{2.97 *10^{8}}[/tex]

      [tex]t = 2.739 *10^{8} \ s[/tex]

Answer:

A compound

Explanation:

A compound is a substance formed when two or more elements are chemically joined

Answer:

Compound

Explanation:

A compound is a substance derived from the chemical combination of two or more elements

e.g Water ;

= [tex]H_2O\\Hydrogen\:and\:Oxygen[/tex]

Salt ;

[tex]NaCl\\Sodium\:and\: Chlorine[/tex]

Answer:

The current needed is 1790.26 A

Explanation:

Given;

magnitude of magnetic field, B = 1.5 T

length of the solenoid, L = 1.8 m

diameter of the solenoid, d = 75 cm = 0.75 m

The magnetic field is given by;

[tex]B = \frac{\mu_o NI }{L}[/tex]

Where;

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

I is current in the solenoid

N is the number of turns, calculated as;

[tex]N = \frac{Length \ of\ solenoid}{diameter \ of \ wire} \\\\N = \frac{1.8}{1.5*10^{-3}} =1200 \ turns[/tex]

The current needed is calculated as;

[tex]I = \frac{BL}{\mu_o N} \\\\I = \frac{1.5 *1.8}{4\pi *10^{-7} *1200} \\\\I = 1790.26 \ A[/tex]

Therefore, the current needed is 1790.26 A.

Answer:

I = 1790.5 A

Explanation:

The magnetic field due to a solenoid is given by the following formula:

B = μ₀NI/L

where,

B = Magnetic Field Required = 1.5 T

μ₀ = 4π x 10⁻⁷ T/A.m

L = length of Solenoid = 1.8 m

I = Current needed = ?

N = No. of turns = L/diameter of wire = 1.8 m/1.5 x 10⁻³ m = 1200

Therefore,

1.5 T = (4π x 10⁻⁷ T/A.m)(1200)(I)/1.8 m

I = (1.5 T)(1.8 m)/(1200)(4π x 10⁻⁷ T/A.m)

I = 1790.5 A

Answer:

Calcium has two electrons in its outer shell.

Explanation:

Calcium is defined as a metal due to its physical and chemical traits. The two outer electrons are very reactive. Calcium has a valence of 2.

Answer:

0.03143 Gy

Explanation:

Mass of the human body = 70 kg

Mass of potassium in the human body = 140 g

chemical atomic mass of potassium = 39.1

From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms

Thus,

140g of potassium will contain;

(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms

We are told that the natural abundance of one of the 40K isotopes is 0.012%.

Thus;

Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms

Formula for activity of K-40 is given as;

Activity = (0.693 × number of K-40 atoms)/half life

Activity = (0.693 × 7.2276 × 10^(19))/1300000000

Activity = 3.85 × 10^(10)

We are told that each decay deposits 1.0 MeV of energy into the body.

Thus;

Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV

Now, 1 MeV = 1.602 × 10^(-13) joules

Thus;

Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J

1 Gy = 1 J/kg

Thus;

Yearly dose = 2.25/70 = 0.03143 Gy

Answer:

A. Ef/ Ei = 1/2

B. EF/ Ei = 1

C Ef / Ei = 4

Explanation:

To solve this we apply Coulomb's law which States that

E = Kq / r^2

Where

q = charge r = straight line distance from q to the point in question and

K = Coulomb's constant

Then

Ei = K Q / r^2

So

A) If Q is halved then

Ef = K Q / (2 r^2)

Ef/Ei = 1/2

B) If R is halved, the value of the E-f

at a distance r remains unchanged. So

Ef/Ei = 1

C) if r is now r/2 then

Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2

Ef / Ei = 4

Answer:

3x10^5m/s

Explanation:

See attached file

Explanation:

The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.

Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.

[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]

Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.

[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]

[tex]v = 3\times 10^5\;\rm m/s[/tex]

Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

To know more about the doppler effect, follow the link given below.

https://brainly.com/question/1330077.

Answer:

Yes, yes it would

Explanation:

Answer:

y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Explanation:

The diffraction of a slit is given by the expressions

          a sin θ = m λ

where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.

          sin θ = m λ / a

If this equation

          a ’= 2 a

we substitute

          2 a sin θ'= m λ

          sin θ'= (m λ / a)  1/2

          sin θ ’= sin θ / 2

We can use trigonometry to find the width

         tan θ = y / L

as the angle is small

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L  

we substitute

        y ’/ L = y/L   1/2

        y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Answer:

velocity and acceleration

Answer:

Hey there!

Centripetal (Circular Motion) and Oscillating Motion.

Let me know if this helps :)

Answer:

The induced current is clockwise

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

Answer:

A) E(r) = 1.3957 × 10^(5) N/C

B) E(r) = 9.8864 × 10⁴ N/C

C) E(r) = 1.13 × 10^(5) N/C

Explanation:

We are given;

q = 6 nc = 6 × 10^(-9) C

L = 10 cm = 0.1 m

d = 4.4 cm = 0.044 m

r1 = 1 cm = 0.01 m

r2 = 2 cm = 0.02 m

r3 = 3 cm = 0.03 m

Formula for the electric field strength in this question is given as;

E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)

When factorized, we have;

E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]

Plugging in the relevant values for q/(2π(ε_o)L)

We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m

Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53

Thus;

E(r) = 1078.52 [1/r + 1/(d - r)]

A) E1 is at r = 1 cm = 0.01m

Thus;

E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))

E(r) = 1.3957 × 10^(5) N/C

B) E2 is at r = 2 cm = 0.02 m

Thus;

E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))

E(r) = 9.8864 × 10⁴ N/C

C) E2 is at r = 3 cm = 0.03 m

Thus;

E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))

E(r) = 1.13 × 10^(5) N/C

Answer:

Increase the distance by a factor of 6.

Explanation:

The intensity at a distance r is given by :

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Here,

P is power emitted

r is distance from source

It means that the intensity is inversely proportional to the distance from the source.

To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, we can increase the distance by a factor of 6. Hence, this is the required solution.

Answer:

The polarity of the magnetic field changes

Explanation:

This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]

[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]

[tex]\alpha=\cos^{-1}(0.3846)[/tex]

[tex]\alpha=67.38^{\circ}[/tex]

We need to calculate the angle β

Using cosine law

[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]

[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]

[tex]\beta=\cos^{-1}(0.923)[/tex]

[tex]\beta=22.63^{\circ}[/tex]

We need to calculate the force on 130 cm side

Using formula of force

[tex]F_{130}=ILB\sin\theta[/tex]

[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]

[tex]F_{130}=0[/tex]

We need to calculate the force on 120 cm side

Using formula of force

[tex]F_{120}=ILB\sin\beta[/tex]

[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]

[tex]F_{120}=0.1385\ N[/tex]

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

[tex]F_{50}=ILB\sin\alpha[/tex]

[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]

[tex]F_{50}=0.1385\ N[/tex]

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.

b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.

c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.

Given the following data:

Let us assume the current is flowing in a counterclockwise direction in the right-angle triangle.

First of all, we would determine the angles by using cosine rule:

[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]

[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]

a. To the determine the magnitude of the magnetic force on the 130 cm side:

Mathematically, the force acting on a current in a magnetic field is given by the formula:

[tex]F = BILsin\theta[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]

b. To the determine the magnitude of the magnetic force on the 120 cm side:

[tex]F_{120}=BILsin\beta[/tex]

[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]

c. To the determine the magnitude of the magnetic force on the 50 cm side:

[tex]F_{50}=BILsin\alpha[/tex]

[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]

Read more: https://brainly.com/question/13754413

Answer:

C. mv to the right

Explanation:

momentum of thr particle=m1v1

momentum of the wall=m2v2

m1v1+ m2v2 =m1u1+ m2u2 since the wall doesn't move it's momentum is zero.

m1v1 =m1u1

therefore change in that occurs as result of the collision is C. mv to the right

The total change in momentum of the particle of mass m that collides elastically with a vertical rigid wall perpendicularly from the left with velocity v is 2mv to the left (option B).  

The total change in momentum is given by:

[tex] \Delta p = p_{f} - p_{i} [/tex]  

In the initial state, the particle is moving to the right until it collides with the rigid wall, so:

[tex] p_{i} = mv [/tex]

In the final state, the particle moves backward after the collision with the wall, so:

[tex] p_{f} = -mv [/tex]  

The minus sign is because it is moving in the negative x-direction (to the left)

Hence, the total change in momentum is:

[tex] \Delta p = -mv - mv = -2mv [/tex]

Therefore, the total change in momentum of the particle is 2mv to the left (option B).

Find more here:

I hope it helps you!

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)

Answer:

The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

[tex]v=\dfrac{c}{\mu}[/tex]

Put the value into the formula

[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]

[tex]v=1.06\times10^{8}\ m/s[/tex]

We need to calculate the time of short pulse of light beam

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]

[tex]t=2.37\times10^{-9}\ sec[/tex]

Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Answer:

51. 7m/s

Explanation:

Take speed of sound in air = 340 m/s

fp = fs (V + Vp)/(V + Vs)

128 = 123 (340 + Vp)/(340 + 36.4)

Vp = 51.7m/s

Explanation:

Answer:

d. 629 nm

Explanation:

slit separation d = .16 x 10⁻³ m

distance of screen D = 1.4 m

distance of second bright band = 11 x 10⁻³

distance of second bright band = 2 x band width

= 2 x λ D /d

Putting the values given ,

11 x 10⁻³  = 2 x λ x 1.4 /  .16 x 10⁻³

λ = 1.76 x 10⁻⁶ / 2.8

= .6285 x 10⁻⁶

= 628.5 x 10⁻⁹

= 629 nm approx .

Answer:

The maximum wavelength of the e-m wave is 6.9 x 10^-7 m

Explanation:

Energy required to trigger a response = 1.8 eV

we convert to energy in Joules.

1 eV = 1.602 x 10^-19 J

1.8 eV = [tex]x[/tex] J

[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J

The energy of an electromagnetic wave is gotten as

E = hf

where

h is the Planck's constant = 6.63 x 10^-34 J-s

and f is the frequency of the wave.

substituting values, we have

2.88 x 10^-19 = 6.63 x 10^-34 x f

f = (2.88 x 10^-19)/(6.63 x 10^-34)

f = 4.34 x 10^14 Hz

We know that the frequency of an e-m wave is given as

f = c/λ

where

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength of the e-m wave

From this we can say that

λ = c/f

λ = (3 x 10^8)/(4.34 x 10^14)

λ = 6.9 x 10^-7 m

Answer:

I. light house 1 will be warmer during the day ii. light house 2 will be warmer at night.

Explanation:

Because the land conducts heat better than water the light house farthest away from the water will get hotter during as the ground will heat up faster than the water. But this also means that the ground will lose heat faster at night where the water won't making the light house closest to the water hotter at night.

Answer:

power = 600000 W

intensity = 6666666.66 W/m²

Em = 70880.18 N/m

F = 2 × [tex]10^{-3}[/tex] N  

Explanation:

given data

frequency f = 2400 MHz

height oven cavity h = 25 cm = 0.25 m

base area measures A =  30 cm by 30 cm

total microwave energy content of cavity E = 0.50 mJ = 0.50 × [tex]10^{-3}[/tex]  

solution

first, we get here total time taken from top to bottom that is express as

Δt = [tex]\frac{h}{c}[/tex]  ...............1

Δt = [tex]\frac{0.25}{3\times 10^8}[/tex]  

Δt = 8.33 × [tex]10^{-10}[/tex] s

and

power output will be

power = [tex]\frac{E}{\Delta t}[/tex]   ..............2

power = [tex]\frac{0.50 \times 10^{-3}}{8.33 \times 10^{-10}}[/tex]

power = 600000 W

and

intensity of the microwave beam is  

intensity = power output ÷ base area    ..............2

intensity =  [tex]\frac{600000}{30 \times 30 \times 10^{-4}}[/tex]  

intensity = 6666666.66 W/m²

and

electric field amplitude is

as we know intensity  I = [tex]\frac{E^2}{c \mu o}[/tex]     ...............3

[tex]E(rms) = \sqrt{Ic\ \mu o} \\E(rms) = \sqrt{6666666.66 \times 3 \times 10^{8} \times 4 \pi \times 10^{-7} }[/tex]

E(rms) = 50119.87 N/m

and we know

[tex]E(rms) = \frac{Em}{\sqrt{2}}\\50119.87 = \frac{Em}{\sqrt{2}}[/tex]  

Em = 70880.18 N/m

and

force on the base due to the radiation is by the radiation pressure

[tex]Pr = \frac{l}{c}[/tex]    ..................4

[tex]\frac{F}{A} = \frac{l}{c}[/tex]  

so

F = [tex]\frac{6666666.66 \times 900 \times 10^{-4}}{3\times 10^8}[/tex]  

F = 2 × [tex]10^{-3}[/tex] N  

Explanation:

We know that rate of flow through a cross section :

[tex]v1 \times a1 = v2 \times a2[/tex]

5 m/s * 1.76cm^2 = v2 * 7.06cm^2

[tex]v2 = 1.24 \: m {s}^{ - 1} [/tex]

At a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

Fluid Flow, a branch of fluid dynamics, is concerned with fluids. It involves the movement of a fluid under the influence of uneven forces. As long as unbalanced pressures are applied, this motion will persist.

Given parameters:

Initial velocity of the water: u = 5.00 m/s

Initial diameter of the pipe: d = 1.50 cm.

Final diameter of the pipe: D = 3.00 cm.

Final velocity of the water: v = ?

In fluid motion:

velocity×(diameter)² = constant

Hence, initial velocity × ( initial diameter)² = final velocity × ( final diameter)²

ud² = vD²

v = u (d/D)²

v= 5 × (1.50/3.0)²

v= 5/2²

v= 5/4

v= 1.25 m/s.

Hence, at a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

Learn more about fluids flow here:

https://brainly.com/question/28727482

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Answer:

Explanation:

Convert all lengths to meters:

Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:

These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].

The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)

Therefore:

[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].

On the other hand:

[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].

Because the cotangent of [tex]\theta[/tex] is very close to zero,

[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].

[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].

[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].

If the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.

On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:

[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].

Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].

Answer:

1.8/0.61 =2.95 ft

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Tysm!

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