For a body to be in SHM, the restoring force must be __________ to displacement from equilibrium.

a) 4.8 m/s² his average acceleration during the first 2.5 s

b) 15 m distance he cover during the first 2.5 seconds.

c) 31 m/s his speed as he finished the race.

Average acceleration is calculated by using Newtons first law of motion

v = u + at

where

u is initial velocity

v is final velocity

a is constant acceleration

t is time

Given u = 0m/s , v = 12 m/s and t = 2.5 sec

Therefore average acceleration is given by

[tex]a=\frac{v}{t}[/tex]

[tex]a=\frac{12}{2.5}[/tex]

a = 4.8 m/sec²

b) Here we will use newtons second law of motion

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

On substituting value we get

s = 15m

c) Here we will use newtons third law of motion

v² = u² + 2as

Here u = 0 , a = 4.8 m/s² and s = 100 m

Therefore

v² = 960

v = 31 m/s

Disclaimer: the question was given incomplete in the portal. Here is the complete question.

Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.

a. What was his average acceleration during the first 2.5 s?

b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?

c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

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The voltage is reduced by a factor of 2.

What is resistance?

                                                     R = V I.

        where,

        R is the resistance of the resistor R in ohms (Ω)

        V is the voltage   drop in the resistor in volts.

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Answer:

C=F×W

so, W=C/F

Explanation:

frequency=F

wavelength=W

speed of light =C in that medium

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

The velocity of the wave is calculated as follows;

v = √T/μ

where;

v = √(50/0.002)

v = 158.1 m/s

f₀ = v/λ

where;

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

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The central pitch around which a scale and its associated harmonies is built is called Tonic.

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_______________

λ=2,4 nm = 2,4·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

_______________

[tex]\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{2,4\cdot 10^{-9}\; m} =517,5\; eV\approx \boldsymbol{518\; eV}[/tex]

To determine whether a celestial body is approaching or receding from Earth, astronomers employ The Doppler principle.

There is a Doppler effect for both light and sound.

For instance, astronomers regularly measure the amount to which a star or galaxy's light is "stretched" towards the lower frequency, red region of the spectrum to calculate how quickly the object is travelling away from us.

A wave's frequency changes in response to an observer moving in respect to the wave source, which is known as the Doppler effect or Doppler shift.

The Doppler effect results from the fact that as the wave source moves in the direction of the observer, the crest of each succeeding wave emerges from a location that is closer to the observer than the crest of the previous wave.

As a result, each wave travels to the observer slightly faster than the one before it. As a result, the interval between successive wave crests arriving at the observer is shorter, increasing the frequency.

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The force that the spring exerts is 1.31 N.

[tex]E_{stored} =\frac{1}{2} kx^{2}[/tex]

F = kx

Where,

x = displacement in the spring

k = spring force constant

Given,

k = 8.75 n/m

x = 0.150 m

Force exerted by the spring, F = kx

putting the values,

F = 8.75 x 0.150

= 1.3125 N

Hence, the force that the spring exerts is 1.31 N.

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They are helium nuclei, which consist of two protons and two neutrons. The net spin on an alpha particle is zero. They result from large, perilous atoms via a process called alpha decay.

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A process in which a large nucleus splits into two or more parts is called ?

(A) nuclear fission

An alpha particle is made up of _____.

(C) two protons and two neutrons

Which statement best describes the type of radioactive decay shown in the model?

(D) The model shows alpha decay, which is a type of nuclear fission.

Is beta decay a form of nuclear fission?

Responses

(A) No, it is not because the mass number does not change.

Which statement correctly classifies gamma emission?

(D) Gamma emission is a type of radioactive decay, but it is not a fission reaction.

i just did the quick check, hope this helps.

The distance covered by the car before it stops is 98.58 m.

Force acting on the car will be

ma = -mg sinθ

where,

m = mass of the car

a = acceleration of the car

g = acceleration due to gravity

θ = angle of inclination

Given,

θ = 10°

Put the values in the above equation,

ma = -mg sinθ

a = -g sin 10°

a = -10 (0.173)

a = -1.73 m/s²

We know the initial velocity is 18.2m/s and when the car comes to a stop the final velocity becomes zero.

u = 18.2 m/s

v= 0 m/s

Using one of the kinematics equations,

v² - u² = 2as

0 - (18.2)² = 2 (-1.73) (s)

331.24 = 3.36 (s)

s = 98.58 m

Hence, the distance covered by the car before it stops is 98.58 m.

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The angle the velocity of the electron make with the magnetic field is     sin(∅) = 90° answer.

  The equation F = qv X B may be damaged down into two elements, the significance and the path. The importance is given by F = qv B sin(∅),

Magnetic field =  1. 25-tesla

                  F = qv B sin(∅)

               = sin(∅) = F/qv B

                 sin(∅) =  qv X B/qv B

                   sin(∅) = 90° answer.

The most importance of the magnetic pressure appearing at the electron whilst its velocity v is perpendicular to B i.e magnetic discipline. The route of the force acting on the electron is the course the palm of the left hand faces. The path of the magnetic area, the direction of the transferring rate, and the route of the pressure on the particle are all perpendicular to each other.

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Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.

Thermo-Electrochemical converter converts potential energy difference during oxidation and reduction of hydrogen in two cell structure to heat energy.

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We will decrease the wavelength of light in order to cause electrons to be ejected from the surface of this metal.

We have a Light of a given wavelength that is used to illuminate the surface of a metal, however, no photoelectrons are emitted.

We have to find out what can we do in order to cause electrons to be ejected from the surface of this metal.

The emission of electrons from the surface of the metal when the light of specific frequency (greater than the threshold frequency) falls over it is called photoelectric effect.

Light consists of photons. The energy associated with the photons is used to emit out the electrons from the surface of metal. We know that - Energy can neither be created nor be destroyed and it can only be transferred from one body to another. Hence, the energy of these moving photons is used to emit electrons from the metal surface. The energy associated with the photon is given by -

E = hμ

Where - μ is the frequency of light an h is Planck's constant.

Now, we can see that the energy of the photon is directly proportional to the frequency of light. The minimum frequency required to eject the electron from the metal surface is called Threshold frequency. Thus, we can emit the electron from the metal surface by using the light of frequency greater than threshold frequency.

Hence, we will increase the frequency of light in order to cause electrons to be ejected from the surface of this metal

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Answer:

well, it is not a solid path and thats a teeny tiny fact

Radio waves in a vacuum travel at the speed of light because they are a type of electromagnetic radiation like a light has been measured as traveling at 3×10^8 m/s in a vacuum.

Charged particles that are accelerating, like time-varying electric currents, are what produce radio waves. Radio and television signals are transmitted using radio waves, and microwaves used in radar and microwave ovens are also radio waves. Radio waves are emitted by a lot of celestial bodies, including pulsars. High RF exposure levels have the potential to heat biological tissue and raise body temperature. The body's inability to handle or remove the extra heat that could be generated by high RF exposure in humans could result in tissue damage.

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The flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

Water flowing in pipes is like flowing of electricity in a circuit. A battery is like a pump from where electricity comes and moves in the circuit. Electrons flowing through wires are like water molecules flowing through pipes. So in comparison between water and electricity, both water and electricity are similar to each other in flowing and movement.

So we can conclude that the flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

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Answer:

focal length

Explanation:

This describes focal length of the beam

The theorem of conservation of mechanical energy is used

a) The initial kinetic energy is, K = 1.33 10⁻⁶ J

b) The fraction of kinetic energy converted to potential energy, U/K = 1.96

The total mechanical energy which is defined as the sum of kinetic energy plus potential energies, is given as

                                            E = K + U

To find the speed of the flea jump when there is no friction,

Initial point

E₁ = K = ½ m v²

Final point.

At the highest height h = 30 cm = 0.30 m because there is no friction

E₂ = U = mgh

Energy is conserved

E₁ = E₂

½ mv² = mgh

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

v = [tex]\sqrt{2(10)(0.4)}[/tex]

v = [tex]\sqrt{4}[/tex] m/s

v = 2 m/s

A) Now we have to find the initial kinetic energy

K = ½ m v²

K = ½ (0.5 × 10⁻⁶) (2)²

K = 1 × 10⁻⁶ J

B) Now we have to find the fraction of energy lost if we take into account the friction of the air

The initial kinetic energy

K = ½ m v²

Now we have to look for the potential energy until h’= 20 cm = 0.20 m

U = mgh'

The fraction of energy lost,

U/K = mgh'/½ mv²

      = 2gh'/v²

      = 2 × 9.8 × 0.2/ 2

      = 1.96

Therefore,

a) The initial kinetic energy is, K = 1 × 10⁻⁶ J

b) The fraction of kinetic energy converted to potential energy, U/K = 1.96

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A process with a negative change in enthalpy and a negative change in entropy will generally be: spontaneous.

Since the Gibbs free energy is a parameter that tells us whether a chemical reaction is spontaneous (Gibbs free energy less than 0) or nonspontaneous (Gibbs free energy greater than 0) in this situation, we can describe it mathematically as:

                                      ΔG = ΔH - TΔS

Therefore, any process with a negative change in enthalpy and a positive change in entropy will be spontaneous. If the enthalpy and the entropy are both negative, the subtraction becomes always negative, for which the Gibbs free energy is also negative.

One of the most crucial thermodynamic functions for the characterization of a system is the Gibbs free energy. It influences results like the voltage of an electrochemical cell and the equilibrium constant for a reversible reaction, among others.

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If the temperature is doubled to 2t, the rate at which it will now radiate is 160 W.

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The circulation around the cylinder is 0.163Js/kg.

To find the answer, we have to know more about the stream flow.

                [tex]v=50m/s\\L=10N/m\\d=1.23kg/m^3[/tex] , where d is the density at the sea level.

                 [tex]L=dvc\\[/tex]

                 [tex]c=\frac{L}{dv} =\frac{10}{1.23*50}=0.163Js/kg[/tex]

Thus, we can conclude that, the circulation around the cylinder is 0.163Js/kg.

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The current must change at a rate of [tex]10^{4}[/tex] A/sec to induce an emf of 1000 volts.

We have a inductor of inductance 100 millihenry.

We have to calculate at what rate the current in the inductor have to change to induce an emf of 1000 Volts.

According to the faraday's law of induced voltage, the voltage induced in the conductor present in the time changing magnetic field is directly proportional to the negative of the rate of change of magnetic flux passing through it per unit time. Mathematically -

ε = [tex]- N\frac{d\phi}{dt}[/tex]

Where -

ε is the induced voltage

N is the number of turns

[tex]\frac{d\phi}{dt}[/tex] represents the rate of change of magnetic flux

Now, in the question it is given that -

L = 100 mH = 0.1 H

ε = 1000 Volts

Now - the voltage across the inductor is given by :

[tex]V = L \frac{di}{dt}[/tex]

[tex]\frac{di}{dt} = \frac{V}{L} = \frac{1000}{0.1} = 10^{4}[/tex] A/sec

Hence, the current must change at a rate of [tex]10^{4}[/tex] A/sec to induce an emf of 1000 volts.

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The magnitude of acceleration of the block is  5.88 m/s²

Pulley are often used in construction sites to raise or lower heavy load. Pulley system is one of the most widely used type of machine. and Block and tackle system of pulleys is the practical system of pulleys in which one or more pulleys are mounted on the same axle, with one continuous rope passing all round the pulleys.

Given that Block 2 has four times the mass of Block 1. Assume a frictionless pulley and that both the rope and pulley are massless and     g = 9.8 m/s2.

That means M2 = 4M1

The acceleration will be toward the M1

To calculate the magnitude of acceleration of the blocks, we will consider the following

T - [tex]m_{1}[/tex]g = [tex]m_{1}[/tex]a ...... (1)

T - [tex]m_{2}[/tex]g = -[tex]m_{2}[/tex]a ..... (2)

Where

Minus equation 2 from equation 1

( [tex]m_{1}[/tex] + [tex]m_{2}[/tex] )a = ( [tex]m_{2}[/tex] - [tex]m_{1}[/tex])g

But [tex]m_{2}[/tex]  =  4[tex]m_{1}[/tex]

Substitute in the equation above.

([tex]m_{1}[/tex] + 4[tex]m_{1}[/tex])a = (4[tex]m_{1}[/tex] - [tex]m_{1}[/tex])g

5[tex]m_{1}[/tex]a = 3[tex]m_{1}[/tex]g

[tex]m_{1}[/tex] will cancel out

5a = 3 x 9.8

a = 29.4/5

a = 5.88 m/s²

Therefore, the magnitude of acceleration of the block is  5.88 m/s²

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The current through the 10 Ω resistor is 0.586A which is connected with a resistor in series combination.

Voltage divides in a Series combination and current divides in a parallel combination.

Let the three resistors joined in parallel be R₁, R₂, R₃

where, R₁= 4 Ω

R₂ = 6 Ω

R₃ = 10 Ω

Given, a battery of 12V and a 2Ω resistor say r is series with the parallel combination.

Equivalent resistance(R) in parallel combination is:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/4 + 1/6 + 1/10

On solving, R = 1.9 Ω

Now, the Equivalent resistance(R) of parallel combination is in series with r = 2Ω

Let Equivalent resistance of Series combination be R'

R' = R + r

R' = 1.9 + 2 Ω

R' = 3.9 Ω

Now let's calculate the voltage drop in the resistor r = 2Ω

v = i × r           where, i is the current in r and v is the voltage drop across r

v = 3.07 × 2

v = 6.14V

Voltage drop, V' across the Equivalent resistance(R) in parallel combination = Total voltage - voltage drop in the resistor r

V' = 12 - 6.14 V

V' = 5.86V

Now, the Voltage drop, V' across the Equivalent resistance(R) in parallel combination is same for all the three resistors R₁, R₂, R₃

So, Voltage is same in a parallel combination.

V' = I × R₃

5.86 = I × 10

I = 0.586A

Hence, The current through the 10 Ω resistor is 0.586A

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No two electrons in an atom or molecule may have the same four electronic quantum numbers, according to the Pauli Exclusion Principle. Only two electrons can fit into an orbital at a time, hence they must have opposing spins.

According to Pauli's Exclusion Principle, no two electrons in the same atom can have values for all four of their quantum numbers that are exactly the same. In other words, two electrons in the same orbital must have opposing spins and no more than two electrons can occupy the same orbital.

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Segment

The space between two waves is known as a segment in an ECG. Beginning at the end of the P wave and ending at the beginning of the QRS complex is the PR segment. Beginning at the end of the QRS wave and ending at the beginning of the T wave is the ST segment.

An electrocardiogram (ECG) is a quick test that can be used to examine the electrical activity and rhythm of your heart. The electrical signals that your heart beats out each time it beats are picked up by sensors that are affixed to your skin.

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32.88grams of (Ag) silver would plate onto the pure electrode if a current of 6. 8A flowed through the cell for 72 minutes.

Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction:

Ag+(aq)+e→Ag(s).

By Using the equation

Q= I×t,    where Q = charge

                           I = current = 6.8 A

                           t= time in seconds= 72×60=4320s

So,

Q=6.8×4320

Q = 29,376C

We see that 1 mole of electrons produces 1 mole of Ag from the half reaction.

1mole of electrons= 96,500C

xmoles of electrons= 29,376C

Cross multiplying;

29,376= 96,500 × x

Now , dividing both sides of the by 96,500

29,376/96,500=x

0.3044 moles of electrons will produce

0.3044×108 = 32.88grams of Ag

Therefore, 32.88grams of (Ag) silver would plate onto the pure electrode if a current of 6. 8A flowed through the cell for 72 minutes.

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The metallic radius (r) of au is 144.2pm

A face-centered cubic unit cell structure is made up of atoms arranged in a cube with six additional whole atoms placed in the centre of each cube face and a fraction of an atom at each of the cube's four corners. Eight additional unit cells share the atoms at the cube's corner.

Atoms in an FCC crystal are in contact along the unit cell's diagonal.          4r = √2 a, where r is the radius of an atom and a is the edge length of a unit cell.

Putting the values in the above expression,we get

r = √2×408 / 4

r = [tex]\frac{1.414*408}{4}[/tex]

r = [tex]\frac{576.9}{4}[/tex]

r = 144.2 pm

Therefore, The metallic radius (r) of au is 144.2pm

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The mass of the electron is [tex]1.72\times 10^{-30}[/tex] Kg.

To calculate the mass of the electron we are using the formula,

[tex]m=\frac{m_{0} }{\sqrt{1-\frac{v^{2} }{c^{2} } } }[/tex]

Here we are given,

[tex]m_0[/tex]=The rest mass of an electron =  [tex]9. 11 \times 10^{-31}[/tex] kg.

v= The velocity of electron =  0. 85c.

c= The velocity of light = c.  (c=[tex]3\times 10^8[/tex] m/s)

We have to found the mass of an electron = m Kg.

Now we substitute the known values, we found that,

[tex]m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-\frac{(0.85c)^{2} }{c^{2} } } }[/tex]

Or,[tex]m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-(\frac{0.85c}{c})^2 } }[/tex]

Or,[tex]m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-({0.85})^2 } }[/tex]

Or, [tex]m=1.72\times 10^{-30}[/tex] Kg.

From the above calculation we can conclude that, The mass of the electron is [tex]1.72\times 10^{-30}[/tex] Kg.

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