A 4.00-Ω resistor, an 8.00-Ω resistor, and a 24.0-Ω resistor are connected together. (a) What is the maximum resistance that can be produced using all three resistors? (b) What is the minimum resistance that can be produced using all three resistors? (c) How would you connect these three resistors to obtain a resistance of 10.0 Ω? (d) How would you connect these three resistors to obtain a resistance of 8.00 Ω?

Answer:

Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.

Instead the physical changes implies that if you can return to the same substance through a reverse process.

Explanation:

A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.

A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.

A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.  

• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.  

• In a chemical change always a chemical reaction takes place.  

• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.  

• In a physical change no new chemical species form.  

• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.  

• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.  

Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.  

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Answer:

The wavelength is  [tex]\lambda = 589 nm[/tex]

Explanation:

From the question we are told that

    The  distance of the mirror shift  is  [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]

      The number of fringe shift is  n =  792

Generally the wavelength producing this fringes is mathematically represented as

               [tex]\lambda = \frac{ 2 * k }{ n }[/tex]

substituting values

              [tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]

             [tex]\lambda = 5.885 *10^{-7} \ m[/tex]

            [tex]\lambda = 589 nm[/tex]

Answer:

Explanation:

From the figure , it is clear that moment of tension is balanced by moment of weight of plate about the line AB which is acting as axis . If W be the weight of plate ,

moment of tension about AB = moment of weight W about line AB

= W x 2.5 cosθ

moment of tension about AB = 2.5 W cosθ

here only variable  is cosθ which changes when θ changes

So, moment of tension about AB varies according to cosθ.

When θ = 0

moment of tension about AB = 2.5 W x cos 0 = 2.5 W . It is the maximum value of moment of tension .

When θ = 90°

moment of tension about AB = 2.5 W cos 90 = 0

moment of tension about AB = 0

So graph of moment of tension about AB will vary according to graph of

cosθ . It has been shown in the file attached .

Answer:

Frequency

Explanation:

When waves travel from one medium to another, it is only the frequency of the wave that remains constant . when a wave is refracted at the boundary between two media, the wave will slow down and its wavelength decreases. The wave usually bends at the interface between the two media. The wavelength and speed of a wave may change at the boundary between two media but its frequency remains the same.

Hence the frequency of light is its only property that remains constant.

Answer:

 Δy = 1 10⁻⁴ m

Explanation:

In double-slit experiments the constructive interference pattern is described by the equation

           d sin θ = m λ

In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle

         tan θ = y / L

as the angles are small,

         tan θ = sin θ / cos θ = sin θ

substituting

         sin θ = y / L

         d y / L = m λ

         y = m λ / d L

let's apply this formula for each wavelength

λ = 450 nm = 450 10⁻⁹ m

m = 5

d = 5.0 mm = 5.0 10⁻³ m

      y₁ = 5 450 10⁻⁹ / (5 10⁻³  1.4)

      y₁ = 3.21 10⁻⁴ m

we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m

      y₂ = 5 590 10⁻⁹ / (5 10⁻³  1.4)

      y₂=  4.21 10⁻⁴ m

the separation of these two lines is

        Δy = y₂ - y₁

        Δy = (4.21 - 3.21) 10⁻⁴ m

        Δy = 1 10⁻⁴ m

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Answer:

The distance of the father from the center is  [tex]d_f = \frac{1}{2} \ m[/tex]

Explanation:

From the question we are told that

    The distance of the son from the center is  [tex]d_s = 2 \ m[/tex]

Let the mass of the son be  [tex]m_s[/tex]

     then the mass of the father is  [tex]m_f = 4m_s[/tex]

Now for the teeter-totter to be balanced the torque due to the weight of the father must be equal to the torque due to the weight the son, this is mathematically represented as

         [tex]\tau_s = \tau_f[/tex]

Where  [tex]\tau_s[/tex] is the torque of the son which is mathematically represented as

        [tex]\tau_ s = m_s * d_s * g[/tex]

while [tex]\tau_f[/tex] is the torque of the father which is mathematically represented as

        [tex]\tau_f = m_f * d_f * g[/tex]

=>    [tex]\tau_f = 4 m_s * d_f * g[/tex]

 So

         [tex]4 m_s * d_f * g = m_s * d_s * g[/tex]

substituting values

        [tex]4 * d_f * = 2[/tex]

 =>    [tex]d_f = \frac{1}{2} \ m[/tex]      

Complete question is;

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

150.42 V

Explanation:

We are told that the circuit consists of four 100W lamps.

We know that Power is given by the equation;

P = V²/R

Thus;

R = V²/P

Now, we are told that the four lamps are connected in parallel across a 230V supply.

Thus, V = 230 V

So resistance, R = 230²/100

R = 529 Ω

We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.

Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω

So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω

Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω

Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω

Thus;

circuit current = 230/2293.5 = 0.100283 A

Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V

CHECK THE COMPLETE QUESTION BELOW

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

the reading of the voltmeter=150.4V

Explanation:

We can determine the wattage of a lamp using below expression:

: W = I² R....................eqn(1)

But fro ohms law V=IR

then I= V/R

If we substitute I into equation (1)

We have W= V²/R

But W= 100W

V= 230V

Then

W=220²/R

100 = 2302/R

R = 529 Ω

We can as well calculate the Resistance of each lamp under given condition that they are sixtimes their value when burning normally.

R = 6 × 529 = 3174 Ω

We can also calculate quivalent resistance of the abovefour lamps connected in parallel then

R = 3174/4

= 793.5 Ω

total circuit resistance can be calculated since we know that resistance is connected to voltmeter of 1500 Ω resistance in series arrangement

Then

total circuit resistance = 1500 + 793.5

= 2293.5 Ω

Then from ohms law again

I= V/R

circuit current = 230/2293.5 A

The reading of the voltage drop across the voltmeter

= 1500 × 230/2293.5

= 150.4V

Answer:

The  angle is  [tex]\theta = 36.24 ^o[/tex]

Explanation:

From the question we are told that

    The  mass is  [tex]m = 0.6 \ kg[/tex]

     The radius is  [tex]r = 1.1 \ m[/tex]

     The speed is  [tex]v = 3.57 \ m /s[/tex]

According to  the law of energy conservation

  The  potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

      [tex]m * g * h = \frac{1}{2} * m * v^2[/tex]

 =>    [tex]h = \frac{1}{2 g } * v^2[/tex]

Here h is the vertical distance traveled by the mass  which is also mathematically represented as

      [tex]h = r * sin (\theta )[/tex]

So

     [tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]

substituting values

     [tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]

     [tex]\theta = 36.24 ^o[/tex]

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Answer

1.07E22 Joules

Explanation;

We know that mass expands by a factor

=>>1/√[1-(v/c)²]

But v= 0.509c

So

1/√(1 - 0.509²)

=>>> 1/√(1 - 0.2591)

= >> 1/√(0.7409) = 1.16

But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy

And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.

So Multiplying by 119.84

Kinetic energy will be 1.07x 10^22 joules

Answer:

Pressure, P = 1250 Pa

Explanation:

Given that,

Weight of a brick, F = 50 N

Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm

We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.

A = 40 cm × 10 cm = 400 cm² = 0.04 m²

So,

Pressure,

[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]

So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.

The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.

The given data in the problem is;

W is the weight of a brick = 50 N

The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm

A is the area,

The area is found as;

A=40 cm × 10 cm = 400 cm² = 0.04 m²

The pressure is the ratio of the force and area

[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]

Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

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Answer:

No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.

Explanation:

If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)

For a body under this type of equilibrium,

ΣF = 0   ...1

where ΣF is the resultant force (total effective force due to all the forces acting on the body)

For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as

ΣF = ma   ...2

where m is the mass of the body

a is the acceleration of the body

Substituting equation 2 into equation 1, we have

0 = ma

therefore,

a = 0

this means that if the resultant force acting on a rigid body is zero, then there won't be any force available to produce acceleration on the body.

Answer:

Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.

Explanation:

As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.

Answer:

They both have the same acceleration

Answer:

     e% = 0.99%   this value is within the 5% tolerance given by the manufacturer

Explanation:

Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by

          e% = | R_nominal - R_measured | / R_nominal 100

where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors

let's apply this formula to our case

R_nominal = 10 kΩ = 10000 Ω

R_measured = 100 99 Ω

        e% = | 10000 - 10099.1 | / 10000 100

        e% = 0.99%

this value is within the 5% tolerance given by the manufacturer

Answer:

E = -8.23 ​​10⁻¹⁷ N / C

Explanation:

In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is

         E = k q₁ q₂ / r²

in this case

q₁ is the charge of the proton and q₂ the charge of the electron

         E = - k e² / a₀²

let's calculate

         E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²

         E = -8.23 ​​10⁻¹⁷ N / C

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

a) The moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m².

b) If the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

Given the data in the question;

Mass of skater; [tex]M = 56.5kg[/tex]

a)

When the skater has his arms pulled inward by assuming they are cylinder of radius; [tex]R = 0.11 m[/tex]

Moment of inertia; [tex]I = \ ?[/tex]

From Parallel axis theorem; Moment of Inertia for a cylindrical body is expressed:

[tex]I = \frac{1}{2}MR^2[/tex]

Where M is the mass and R is the radius

We substitute our given values into the equation

[tex]I = \frac{1}{2}\ *\ 56.5kg\ *\ (0.11m)^2\\\\I = \frac{1}{2}\ *\ 56.5kg\ *\ 0.0121m^2\\\\I = 0.3418kg.m^2[/tex]

Therefore, the moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m²

b)

With the skater's arms extended by assuming that each arm is 5% of the mass of their body

Mass of each arm; [tex]M_a = \frac{5}{100} * M = \frac{5}{100} * 56.5kg = 2.825kg[/tex]

Remaining mass; [tex]M_b = M - 2M_a = 56.5kg - 2(2.825kg) = 50.85kg[/tex]

Assume the body is a cylinder of the same size and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.

Length of arm; [tex]L = 0.875 m[/tex]

From Parallel axis theorem; Moment of Inertia about vertical axis is expressed as:

[tex]I = \frac{1}{2}M_bR^2 + \frac{2}{3}M_aL^2[/tex]

We substitute in our values

[tex]I = \frac{1}{2}*50.85kg*(0.11m)^2 + \frac{2}{3}*2.825kg*(0.875m)^2\\\\I = [\frac{1}{2}*50.85kg * 0.0121m^2] + [\frac{2}{3}*2.825kg*0.765625m^2]\\\\I = 0.3076kg.m^2 + 1.4419kg.m^2\\\\I = 1.7495kg.m^2[/tex]

Therefore, if the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

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Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

area of the square coil, A = 0.4 m²

number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

E = -NAB/t

E = -(15 x 0.4 x 0.75) / 0.05

E = -90 V

|E| = 90 V

Therefore, the magnitude of the average induced emf is 90V

Answer:

[tex]\boxed{\sf 65.3 \ inches}[/tex]

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Answer:

In an adiabatic process we have

pV γ = const..

This explains that the pressure is a function of volume, p ( V ) ,

So can be written as:

p ( V ) × V γ = p 0 V γ 0 ,

or p ( V ) = p 0 V 0 / V γ

= p 0 V 0 / V ^(7 / 5)

Answer:

None of the above

Explanation:

The formula of the magnetic field of a point next to a wire with current is:

B = 2×10^(-7) × ( I /d)

I is the intensity of the current.

d is the distance between the wire and the point.

● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T

Answer:

A

 [tex]\lambda = 361.45 \ m[/tex]

B

[tex]k = 0.01739 \ rad/m[/tex]

C

 [tex]w = 5.22 *10^{6} \ rad/s[/tex]

D

[tex]E = 0.01446 \ N/C[/tex]

Explanation:

From the question we are told that

   The frequency is [tex]f = 83 0 \ kHz = 830 *10^{3} \ Hz[/tex]

    The  magnetic field amplitude is  [tex]B = 4.82*10^{-11} \ T[/tex]

Generally wavelength is mathematically represented as

        [tex]\lambda = \frac{c}{f}[/tex]

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

     =>  [tex]\lambda = \frac{3.0*10^{8}}{ 830 *10^{3}}[/tex]

     =>  [tex]\lambda = 361.45 \ m[/tex]

Generally the wave number is mathematically represented as

        [tex]k = \frac{2 \pi }{\lambda }[/tex]

=>     [tex]k = \frac{2 * 3.142 }{ 361.45 }[/tex]

=>    [tex]k = 0.01739 \ rad/m[/tex]

Generally the angular frequency is mathematically represented as

     [tex]w = 2 * \pi * f[/tex]

=>  [tex]w = 2 * 3.142 * 830*10^{3}[/tex]

=>   [tex]w = 5.22 *10^{6} \ rad/s[/tex]

The the electric-field amplitude is mathematically represented as

     [tex]E = B * c[/tex]

=>    [tex]E = 4.82 *10^{-11} * 3.0*10^{8}[/tex]

=>     [tex]E = 0.01446 \ N/C[/tex]

This question involves the concepts of wavelength, frequency, wave number, and electric field.

a) The wavelength is "361.44 m".

b) The wave number is "0.0028 m⁻¹".

c) The angular frequency is "5.22 x 10⁶ rad/s".

d) The electric field amplitude is "0.0145 N/C".

a)

The wavelength can be given by the following formula:

[tex]c=f\lambda[/tex]

where,

c = speed of light = 3 x 10⁸ m/s

f = frequency = 830 KHz = 8.3 x 10⁵ Hz

λ = wavelength = ?

Therefore,

[tex]3\ x\ 10^8\ m/s=(8.3\ x\ 10^5\ Hz)\lambda\\\\\lambda=\frac{3\ x\ 10^8\ m/s}{8.3\ x\ 10^5\ Hz}\\\\[/tex]

λ = 361.44 m

b)

The wave number can be given by the following formula:

[tex]wave\ number = \frac{1}{\lambda} = \frac{1}{361.44\ m}[/tex]

wave number = 0.0028 m⁻¹

c)

The angular frequency is given as follows:

[tex]\omega = 2\pi f = (2)(\pi)(8.3\ x\ 10^5\ Hz)[/tex]

ω = 5.22 x 10⁶ rad/s

d)

The electric field amplitude can be given by the following formula:

[tex]\frac{E}{B} = c\\\\c(B)=E\\\\E = (3\ x\ 10^8\ m/s)(4.82\ x\ 10^{-11}\ T)\\[/tex]

E = 0.0145 N/C

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Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}[/tex]

But heat capacity of object B is twice that of object A

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K[/tex]

Therefore, the final temperature of both objects is 400 K.

Explanation:

Using Equations of Motion :

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

Height = 0 * 4 + 4.9 * 16

Height = 78.4 m

Answer:

The potential will be Va/b

Explanation:

So Let sphere A charged Q to potential V.

so, V= KQ/a. ....(1

Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.

therefore, potential will be ,

V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]

Answer:

The emerging intensity is equal to 0.75[tex]I_{o}[/tex]

Explanation:

The initial intensity of the light = [tex]I_{o}[/tex]

The angle of polarization β = 30°

We know that the polarized light intensity is related to the initial light intensity by

[tex]I[/tex] = [tex]I_{0} cos^{2}\beta[/tex]

where [tex]I[/tex] is the emerging polarized light intensity

inserting values gives

[tex]I[/tex] = [tex]I_{0} cos^{2}[/tex] 30°

[tex]cos^{2}[/tex] 30° = [tex](cos 30)^{2}[/tex] = [tex](\frac{\sqrt{3} }{2} )^{2}[/tex] = 0.75

[tex]I[/tex] = 0.75[tex]I_{o}[/tex]

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]

So,

[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.