Which of the units of the following physical quantities are derived

Answer:

5.2791264*10¹³

Explanation:

Convert the 9 years to seconds and then multiple it by 186000

The star is 4.62 x 10¹⁶ miles away from Earth.

The speed of light is 186,000 miles per second. It takes light from a particular star approximately 9 years to reach Earth. There are 365 days in 1 year, so it takes 9 x 365 = 3285 days for light from the star to reach Earth.

The distance between the star and Earth is 3285 x 186,000 = 608,810,000 miles. In scientific notation, this is 4.62 x 10¹⁶ miles.

Here is the calculation:

distance = speed * time

distance = 186,000 miles/second * 3285 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute

distance = 608,810,000 miles

distance = 4.62 x 10¹⁶ miles

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Answer:

Explanation:

1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;

M ∝ V

M = kV

For every proportionality sign, there will always be a proportionality constant 'k'

Since the proportionality constant is the density (D) of the metal, the equation will become;

M = DV

Given the density to be 11.3 g/cm3, the equation will become;

M = 11.3V

Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V

2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;

M = 11.3V

M = 11.3 * 17.3

M = 195.49 grams

Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.

Explanation:

[tex]v = \sqrt{2ax}[/tex]

Take the square of both sides:

[tex]v^2 = \left(\sqrt{2ax}\right)^2 = 2ax[/tex]

Divide both sides by 2a and you will get

[tex]x = \dfrac{v^2}{2a}[/tex]

Answer:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

Answer:

 f = - 20 cm

Explanation:

This exercise asks us for the focal length, which for a lens in air is

                  1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)

where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.

In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression

                 - 1/40 = (n-1) (1 / R₁ -1 / R₂)

                (1 / R₁ - 1 / R₂) = - 1/40 (n-1)

let's calculate

               (1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)

               (1 / R₁ -1 / R₂) = -1/20

 Now we change the construction material for one with refractive index

n = 2, keeping the radii,

              1 / f = (n-1) (1 / R₁-1 / R₂)

              1 / f = (n-1) (-1/20)

let's calculate

             1 / f = (2.00-1) (-1/20)

              1 / f = -1/20

              f = - 20 cm

Answer:

The wavelength is  [tex]\lambda = 1805 nm[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]

     The  distance of the screen is  D  =  3.0  m

     The  fringe width is  [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]

Generally the fringe width for a bright fringe  is mathematically represented as

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

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

=>     [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]

=>     [tex]d = 0.000373 \ m[/tex]

Generally the fringe width for a dark fringe  is mathematically represented as

      [tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

Here  m = 0  for  first order dark fringe

   So  

         [tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

looking at which we see that   [tex]y_d = y[/tex]

         [tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]

=>    [tex]\lambda = 1805 *10^{-9} \ m[/tex]

=>    [tex]\lambda = 1805 nm[/tex]

Answer:

4 x 10¹⁵

Explanation:

Answer and Explanation:

Data provided as per the question is as follows

Speed at point A = 20 m/s

Acceleration at point C = [tex]5 m/s^2[/tex]

[tex]r_A = 25 m[/tex]

The calculation of the magnitude of the acceleration at A is shown below:-

Centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

now we will put the values into the above formula

= [tex]\frac{20^2}{25}[/tex]

After solving the above equation we will get

[tex]= 16 m/s^2[/tex]

Tangential acceleration is

[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]

Answer:

v = - 14.08 m / s

Explanation:

The definition of center of mass is

        [tex]x_{cm}[/tex] = 1 /M  ∑sun [tex]x_{i} m_{i}[/tex]

where M is the total mass of the system and [tex]x_{i}[/tex] and [tex]m_{i}[/tex] are the position and mass of each component.

The velocity of the center of mass can be found by deriving this expression with respect to time

         [tex]v_{cm}[/tex] = 1 / M ∑ m_{i} [tex]v_{i}[/tex] vi

let's find the total mass

          M = m₁ + m₂ + m₃ + m₄

          M = 1.45 + 2.81 +3.89 + 5.03

          m = 13.18 kg

let us substitute in the velocity of the center of mass [tex]v_{cm}[/tex] = 0

          0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)

          v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄

let's substitute the given values

v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03

They ask us for the calculations for a time t = 2.83 s

          v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03

          v = - 14.08 m / s

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

[tex]m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg[/tex]

The velocity of the center mass of the particles is calculated as;

[tex]M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}[/tex]

The velocity of particle 1 at time, t = 2.83 s;

[tex]v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s[/tex]

The velocity of particle 2 at time, t = 2.83 s;

[tex]v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s[/tex]

The velocity of particle 3 at time, t = 2.83 s;

[tex]v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s[/tex]

The velocity of the particle 4 at time, t = 2.83 s;

[tex]v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s[/tex]

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

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Answer:

d) Microwave ovens emit in the same frequency band as some wireless Internet devices.

Explanation:

Microwave are radio waves of short wavelength, from about 10 centimetres to one millimetre, in the Super High Frequency and the Extremely High Frequency bands. Microwaves can penetrate into materials and deposit their energy below the surface which is why is is used in microwave heating found in microwave oven. Transmission of data sometimes involves the use of microwaves to send and receive information over a long distance. Microwaves are the mainly used in radar, used for satellite communication, and wireless networking technologies such as Wi-Fi.

Answer:

Explanation:

Equation of refraction of a lens is expression according to the formula given below;

[tex]\dfrac{n_2}{v} = \dfrac{n_1}{u}= \dfrac{n_2-n_1}{R}[/tex]

R is the radius of curvature of the convex refracting surface = 12cm

v is the image distance from the refracting surface

u  is the object distance from the refracting surface

n₁ and n₂ are the refractive indices of air and the medium respectively

Given parameters

R = 12 cm

u = [tex]\infty[/tex] (since light incident is parallel to the axis)

n₁  = 1

n₂  = 1.5

Required

focus point of the light that is incident and parallel to the central axis (v)

Substituting this values into the given formula we will have;

[tex]\dfrac{1.5}{v} - \dfrac{1}{\infty}= \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0= \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}= \dfrac{0.5}{12}\\\\[/tex]

Cross multiply

[tex]1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm[/tex]

Hence  Light incident parallel to the central axis is focused at a point 36cm from the surface

Explanation:

Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]

[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]

[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]

The work function [tex]\phi[/tex] is then

[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]

[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]

The work function of the surface is equal to 3.98 × 10⁻¹⁹J.

The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.

The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.

The expression between wavelength (λ), frequency, and speed of light (c) is:

c = νλ

Given, the wavelength of the light, ν = 500 nm

The frequency of the light can determine from the above-mentioned relationship:

ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz

The work function = h ν =  6 × 10¹⁴ × 6.626 × 10⁻³⁴

φ = 3.98 × 10⁻¹⁹J

Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.

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Answer:

Explanation:

Your body weighs more at the pole for two important reasons. Both have to do to the spin of the earth on its axis.

Because of its spin the earth is thicker around the equator than it is through the poles. This means that when you stand on the equator, you are farther away from the center of earth than you would be at the poles. As gravity decreases with the inverse of the square of distance, gravity will be weaker at the equator.

As you are also spinning with the earth, you will have a required centripetal acceleration and force to keep you attached to the ground, This force decreases the effect of gravity so again, you would weigh less at the equator.

Answer:

The acceleration in the block is 2.1 m/s²

Explanation:

Given that,

Mass = 50 kg

Angle = 54°

Force = 40 N

Coefficient of friction = 0.33

We need to calculate the acceleration in the block

Using balance equation

[tex]F_{net}=F_{f}-F\cos\theta[/tex]

[tex]ma=\mu mg\sin\theta-F\cos\theta[/tex]

[tex]a=\dfrac{\mu mg\sin\theta-F\cos\theta}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{0.33\times50\times9.8\sin54-40\cos54}{50}[/tex]

[tex]a=2.1\ m/s^2[/tex]

Hence, The acceleration in the block is 2.1 m/s²

Explanation:

Let [tex]R_1[/tex] and [tex]R_2[/tex] be the the resistances of the resistors. We are given that

[tex]R_1 + R_2 = 690\:Ω\:\:\:\:\:\:\:(1)[/tex]

and

[tex]\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{118\:Ω}\:\:\:\:\:(2)[/tex]

From Eqn(1), we can write

[tex]R_2 = 690\:Ω - R_1[/tex]

and then plug this into Eqn(2):

[tex]\dfrac{1}{R_1} + \dfrac{1}{690\:Ω - R_1} = \dfrac{1}{118\:Ω}[/tex]

or

[tex]\dfrac{690\:Ω}{(690\:Ω)R_1 - R_1^2}= \dfrac{1}{118\:Ω}[/tex]

[tex]\Rightarrow R_1^2 - (690\:Ω)R_1 + (690\:Ω)(118\:Ω)= 0[/tex]

or

[tex]R_1^2 - 690R_1 + 81420 = 0[/tex]

Using the quadratic formula, we find that the above equation has two roots:

[tex]R_1 = 151.1\:Ω,\:\:538.9\:Ω[/tex]

This means that if you choose one root value for [tex]R_1[/tex], the other root will be the value for [tex]R_2[/tex].

Answer:

Given that

capacitor being charged by a current i c has a displacementcurrent equal to i c between the plates∴

displacement current iD =i c=0.280 Aradius of the circular plate r = 4 cm=0.04 m

( A ) . displacement current density j D = iD / ( π r 2 )=0.28 / ( 3.14 * 0.04 2 )=55.73 A / m 2

( B ) . displacement current density j

D = ε( dE / dt )the rate at which the electric field between the plates is changing(

dE / dt ) = jD/ εdE/ dt ) = 55.73/ 8.85 * 10 -12=6.3*10 12 N / C - s( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R 2) * i c ----( 1 )whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 -7 H( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),wegetanswer

Yeah

All they are all correct

Given that,

Energy [tex]H=2.7\times10^{31}\ W[/tex]

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

[tex]H=Ae\sigma T^4[/tex]

[tex]A=\dfrac{H}{e\sigma T^4}[/tex]

[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]

[tex]R=5.0\times10^{10}\ m[/tex]

(b). Given that,

Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]

[tex]R=5.42\times10^{6}\ m[/tex]

Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]

(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]

Answer:

0.09N, attractive

Explanation:

It can be deducted from the question that the currents are arranged in parallel settings, then it is obvious that the force on each of the wire will be attractive toward the other wire.

the magnitude of force can be determined by using below formula;

F2 = (μ₀/2π)(I₁I₂/d)I₂

μ₀ = constant = 4π × 10^-7 H/m,

I₁, I₂ = currents= 30A

L = the length o the wire=30m

d = distance between these two wires= 0.06m

Since the current are arranged in the same direction, they exhibit attractive force on each other.

Then plugging the values Into the formula above we have

F₂ = (4π × 10^-7 T.m/A)/2π) × ((30A)²/ 0.06m)× 30 m

= 0.09 N, attractive

Therefore, the magnitude and direction of the force is 0.09 N, attractive

Answer:

Explanation:

The work done by brake friction changes the kinetic energy,

ASSUMING the road is level and other friction is ignored.

W = KEf - KEi = ½mvf² - ½mvi² = ½m(vi² - vf²)

W = ½(1800)(30² + 10²) = 900(900 - 100) = 720000 J = 720 KJ

Answer:

An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will __LOSE___ altitude if the altimeter is not corrected

Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.

Answer:

Explanation:

Answer:

strong nuclear is the stronger force; it is related to mass

Explanation:

If we compare the force of gravity to strong nuclear force, we could conclude that strong nuclear is the stronger force; it is related to mass, therefore the correct answer is option C

The nuclear force is the interaction between the subatomic particles that make up a nucleus. There are two types of nuclear forces: the strong nuclear force and the weak nuclear force. Depending on the separation between the proton neutron and proton pairs, these nuclear forces can be both attracting and positive.

Both types of nuclear forces come under the four fundamental forces of nature. There are mainly four fundamental forces of nature electromagnetic force, gravitational force, strong nuclear force, and weak nuclear force.

Thus, Option C is the appropriate response since, when compared to the force of gravity, the strong nuclear force is the greater force because it is tied to mass.

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Answer:

Explanation:

Given data

voltage supplied Vs= 1.5 volts

resistance R1= 1000 ohms

resistance R2= 500 ohms

The total resistance is

Rt= 1000+ 500

Rt= 1500 ohms

The current I is given as

[tex]I= \frac{Vs}{Rt} \\\\ I= \frac{1.5}{1500} = 0.001mA[/tex]

Voltage across R1

[tex]VR1= Vs(\frac{R1}{R1+R2} )=1.5(\frac{1000}{1000+500} )= 1.5(\frac{1000}{1500} )\\ \\\ VR1= 1v[/tex]

Voltage across R2

[tex]VR2= Vs(\frac{R2}{R1+R2} )=1.5(\frac{500}{1000+500} )= 1.5(\frac{500}{1500} ) \\\ VR2=0.5v[/tex]

In series connection the current is the same for all components while the voltage divides across all components,the voltages consumed by each individual resistance is equal to the source voltage.

Answer:

Explanation:

Let f₀ be the frequency .

energy of photons having frequency of f₀

= hf₀ where h is plank's constant

for electron to get ejected , work function should be equal to energy of photon

hf₀ = 3.55

4.1357 x 10⁻¹⁵ x f₀ = 3.55

f₀ = 8.58 x 10¹⁴ Hz .

Complete Question

(c) It takes you 5 hours to to bring the turkey from 10.0°C to 75.0 °C. During that time, the electrical grid transfers a constant 2500.0 Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings?

Answer:

[tex]Q=4.50 *10^7J[/tex]

Explanation:

From the question we are told that:

Time [tex]t=5hours[/tex]

Temperature rise [tex]dT= 65\textdegree[/tex]

Power [tex]P=2500.0 Watts[/tex]

Generally, the equation for Power is mathematically given by

[tex]P=\frac{Q}{t}[/tex]

Therefore

[tex]Q=2500*5*360[/tex]

[tex]Q=4.50 *10^7J[/tex]

Answer:

As opposed to n-type semiconductors, p-type semiconductors have a larger hole concentration than electron concentration.

Explanation:

In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. A common p-type dopant for silicon is boron or gallium. hope this you :)

Answer:

Here are notes I took on semiconductor conductivity :

________________________________________________________
-A p-type semiconductor is made of a material in which electrical conduction is due to the movement of a positive charge.

-Examples of p-type dopants - boron, aluminum, gallium, indium, and thallium

Explanation:

In this case, indium only correct option being a dopant of a P-type semiconductor. Other options are N-type dopants.

Hopefully its correct !! <3

The question is not complete so i have attached it.

Answer:

The light source is 2 cm from the bottom of the lamp

Explanation:

From the attached image, we can see that the parabola opens up with its vertex at the origin.

Now, the standard form of equation for a parabola is:

x² = 4ay

Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)

Thus, we have;

12² = 4a(18)

144 = 72a

a = 144/72

a = 2

Looking at the parabola again, the line of symmetry is at x = 0

Thus, axis of symmetry is at x = 0.

Thus, focus is at (0, 2)

So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm

The distance of the light source from the bottom of the lamp is 2 cm.

The given parameters;

Apply standard parabola equation to determine the distance of the light source from the bottom of the lamp;

[tex]x^2 = 4ay\\\\12^2 = 4a(18)\\\\144 = 72 a\\\\a = \frac{144}{72} \\\\a = 2 \ cm[/tex]

Thus, the distance of the light source from the bottom of the lamp is 2 cm.

"Your question is not complete, it seems to be missing the following information";

the top right edge of the lamp has a coordinate of (12,18)

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