Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s.A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 6.00 Hz, which increases to 7.00 Hz when she tightens her violin string slightly.
(a) What was the frequency of the note played by her violin when she heard the 3-Hz beats?
(b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?

Explanation:

It is given that,

Wavelength of x-rays = 2 nm

Plane spacing, d = 0.281 nm

It is assumed to find the scattering angle for second order maxima.

For 2nd order, Bragg's law is given by :

[tex]2d\sin\theta=n\lambda[/tex]

For second order, n = 2

[tex]\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)[/tex]

Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.

Answer:

1794.47 radian.

Explanation:

Suppose a flagellum is rotating at a constant 952 rps (revolution per second). What is the angular displacement of the flagellum in 0.3seconds? (report your answer in SI units)

Let us suppose that a flagellum is rotating at a constant 952 revolution per second.

Firstly, we will convert 952 revolution per second to radian per second as follows :

952 revolution per second = 2π × 952 rad/s

=5981.592 rad/s

The angular displacement is given by the expression as follows :

[tex]\theta=\omega t\\\\\theta=5981.592\times 0.3\\\\\theta=1794.47\ \text{rad}[/tex]

So, the angular displacement of the flagellum in 0.3 seconds is 1794.47 radian.

Answer:

Explanation:

The minimum magnitude of acceleration = 3 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 3 x 1²

= - 3 + 1.5

= - 1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The maximum  magnitude of acceleration = 6 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 6 x 1²

= - 3 + 3

=  0

position at t = 1 s

= 10 +0

= 10  m

So range of position is 8.5 m to 10 m .

We want to find the range of possible positions for the cart at t = 1s.

The range is:

14.5m ≤ p ≤ 16m

First, we know that the acceleration of the cart is a, a constant (but we don't know the exact value yet) so we write the acceleration as:

a(t) = a

To get the velocity equation we integrate over time, and we know that the velocity at t = 0s is 3m/s, so that will be the constant of integration.

v(t) = a*t + 3m/s

To get the position we integrate again, the initial position is x = 10m, so that will be the constant of integration.

p(t) = (a/2)*t^2 + (3m/s)*t + 10m

Now, to get the range of possible positions for t = 1s we need to use the minimum and maximum accelerations and evaluate the position equation in t = 1s.

The minimum acceleration is a = 3m/s^2, so we have the minimum position:

p(1s) = (3m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 14.5m

The maximum acceleration is 6m/s^2, so the maximum position is:

P(1s) =  (6m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 16m

So the range of the position at t = 1s is:

14.5m ≤ p ≤ 16m

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Answer:

__________________

I think the answer is A.

__________________

Answer:

m = 312.5

Explanation:

Given that,

The focal length of the objecive lens, [tex]f_o=1\ cm[/tex]

The focal length of eye piece, [tex]f_e=2\ cm[/tex]

length of the tube, L = 25 cm

We need to find the magnitude of the overall magnification of the microscope. It is given by the formula as follows :

[tex]m=\dfrac{L}{f_o}\times \dfrac{D}{f_e}[/tex]

D = 25 cm

So,

[tex]m=\dfrac{25}{1}\times \dfrac{25}{2}\\\\m=312.5[/tex]

So, the overall magnification of the microscope is 312.5.

Answer:

a

  [tex]F = 2.32*10^{-17} \ N[/tex]

b

[tex]n =3869 \ electrons[/tex]

Explanation:

From the question we are told that

   The  charge on each water drop is  [tex]q_1=q_2=q = - 6.19*10^{-16} \ C[/tex]

   The  distance of separation is  [tex]d = 1.22\ cm = 0.0122 \ m[/tex]  

Generally the electrostatic force between the water drops is mathematically represented as

      [tex]F = \frac{k * q_1 * q_2 }{ d^ 2}[/tex]

Here  k is the coulombs constant with value  [tex]k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

So  

      [tex]F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}[/tex]

       [tex]F = 2.32*10^{-17} \ N[/tex]

Generally the quantity of charge is mathematically represented as

        [tex]q = n * e[/tex]

Here n is the number of electron present

and  e is the charge on one electron with value  [tex]e = 1.60*10^{-19} \ C[/tex]

    So

         [tex]n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}[/tex]

         [tex]n =3869 \ electrons[/tex]

Answer:

in scientific notation 26500 is 2.65×10⁴

Answer:

2.56*10^4

26500

(We have to move the decimal 4 places)

2.6500*10^4

So the answer is 2.6500*10^4 or 2.56*10^4

Hope this helps! :)

Answer:

a) A Helium nucleus with two protons and one neutron

b) 0.324 of the original amount

Explanation:

a) When Tritium decays, the nucleus emits an electron and an antineutrino, which then changes it from a Triton with one proton and two neutrons, to a Helium nucleus with two protons and one neutron.

The decay constant for the decay of Tritium is gotten from

[tex]t_{1/2}[/tex] = 0.693/k

where

[tex]t_{1/2}[/tex] is the half life = 12.3 yrs

k is the decay constant = ?

substituting, we have

12.3 = 0.693/k

k = 0.693/12.3 = 0.0563

The fraction that will remain after 20 years can be calculated from

[tex]N[/tex] = [tex]N_{0} e^{-kt}[/tex]

where

[tex]N[/tex] is the final remaining amount of Tritium after 20 years

[tex]N_{0}[/tex] is the original amount of Tritium

k is the decay constant = 0.0563

t is the time = 20 years

Equation can be re-written as

[tex]N/N_{0}[/tex] = [tex]e^{-kt}[/tex]

where

[tex]N/N_{0}[/tex] is the fraction of the tritium that will be remaining after 20 years.

substituting values, we have

[tex]N/N_{0}[/tex] = [tex]e^{-0.0563 * 20}[/tex]

[tex]N/N_{0}[/tex]  =  [tex]e^{-1.126}[/tex]

[tex]N/N_{0}[/tex] = 0.324 of the original amount

Answer:

A

Explanation:

Gravitational force given by

F_ps =G(m_pM_s)/r^2

G= gravitational constant

m_p=mass of the probe =125 Kg

M_s= mass of sun

r= distance between probe and Sun.

Similarly, F_pe = G(m_pM_e)/r^2.

m_p = mass of probe

M_e = mass of earth

Clearly, M_s> M_e . Therefore, F_ps> F_pe.

Hence,  A. The force is toward the sun.

Answer:

5.68*10^-5 °C

Explanation:

ADC is an acronym that means analog to digital conversion.

The numbers of bit in any ADC is the level of Voltage it possess.

The question states 32 bits, this means that it's total voltage levels is 2^32

Also, the question gives us a range of 10 V for the ADC. Now we can evaluate the resolution of the voltage as:

Δv = 10 / (2^32)

Δv = 10 / 4.29^9

Δv = 2.33*10^-9

The thermocouple sensitivity is 41 micro volts/°C

Thermocouple sensitivity = Δv/ΔT, where ΔT is the temperature change

ΔT = Δv/thermocouple sensitivity

ΔT = 2.33*10^-9 / 41*10^-6

ΔT = 5.68*10^-5 °C

Thus, the smallest temperature change is 5.68*10^-5 °C

Explanation:

Given:

v₀ = 0 m/s

a = 2.50 m/s²

t = 4 s

Find: v

v = at + v₀

v = (2.50 m/s²) (4 s) + 0 m/s

v = 10 m/s

The speed of the ball after 4s is 10 m/s.

There's not enough information to find its velocity. (We would need to know what direction it's moving.)

Answer:

The new period is 16 s

Explanation:

The period of a simple pendulum is given by

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]

Where [tex]T[/tex] is the period

[tex]L[/tex] is the length of the string

[tex]g[/tex] is the acceleration due to gravity

2π is constant

For the first experiment,

[tex]T =8 s[/tex]

[tex]g = gE[/tex]

Then,

[tex]8 = 2\pi \sqrt{\frac{L}{gE} } \\[/tex]

[tex]8\sqrt{gE} = 2\pi \sqrt{L}[/tex] ....... (1)

For the second experiment

[tex]T = T_{2}[/tex]

[tex]g = \frac{1}{4} gE[/tex]

Hence,

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex] becomes

[tex]T_{2} =2\pi \sqrt{\frac{L}{\frac{1}{4}gE } }[/tex]

Then,

[tex]T_{2} \sqrt{\frac{1}{4}gE } = 2\pi\sqrt{L}[/tex] ....... (2)

Since 2π is constant and

The same pendulum is used for the second experiment, then [tex]L[/tex] is also constant.

∴ [tex]2\pi \sqrt{L}[/tex] is constant for both experiment.

Hence, we can equate equations (1) and (2) such that

[tex]8\sqrt{gE} = T_{2} \sqrt{\frac{1}{4}gE }[/tex]

Then,

[tex]\frac{8\sqrt{gE} }{\sqrt{\frac{1}{4}gE } } = T_{2} \\\frac{8\sqrt{gE} }{\frac{1}{2}\sqrt{gE} } } } = T_{2} \\ 16 = T_{2} \\T_{2} = 16 s[/tex]

Hence, the new period is 16 s

A simple pendulum swings for 8 seconds. The new period is determined by repeating the pendulum experiment on a plant with g = 1/4 gE.

Move to a position where the gravitational acceleration is greater.

- Reduce the pendulum's length.

Explanation: The frequency of a simple pendulum is given by where L is the length of the pendulum g is the gravitational acceleration

From the formula, we notice that - As the value of g increases, the frequency of the pendulum increases as well

- As the value of L increases, the frequency of the pendulum decreases

So, in order to increase the frequency of the pendulum, we can: - Move to a location where the acceleration due to gravity, g, is larger
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Answer:

entropy

Explanation:

GPG is an acronym for GNU Privacy Guard, which is considered as free software, is a command-line tool for generating cryptographically secure keys.

These cryptographically secure keys can be created by using entropy, through reliable entropy source, like /dev/random, which in turn derives its entropy from physical devices connected to the machine, such as keyboards, mice, and disks.

Hence, the right answer is ENTROPY

"Acceleration" means the rate at which velocity is changing.

If velocity is constant, then it isn't changing.

If velocity isn't changing, then acceleration is zero.

Answer:

calculate 1.6*7.8*3.1

Explanation:

Answer:

The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

[tex]E_{i}=E_{f}[/tex]

[tex]\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2[/tex]

[tex]\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})[/tex]

[tex]0.00075(400-v_{f}^2)=0.18 [/tex]

[tex]400-v_{f}^2=\dfrac{0.18}{0.00075}[/tex]

[tex]-v_{f}^2=240-400[/tex]

[tex]v_{f}^2=160[/tex]

[tex]v_{f}=4\sqrt{10}\ m/s[/tex]

Hence, The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]

"Should wolves be reintroduced into national parks" is a question that cannot be answered through making measurements, therefore the correct answer is option C.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

A length, for instance, is a physical quantity. A defined, predetermined length is represented by the length unit known as the meter.

A qualitative type of question cannot be answered while making a quantitative measurement.

Thus, the question of whether wolves should be reintroduced into national parks cannot be answered by taking measurements, so option C is the appropriate response.

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Answer:

x=L/2  y=0 and the charge q3 is ¼ of the charge q

Explanation:

For this exercise we will use Coulomb's law.

         F₁₂ = k q₁ q₂ / r₁₂²

From this expression we see that like charges repel and charges of different signs attract.

Let's apply this expression to our case, they indicate that the two charges are of equal magnitude and sign, therefore the force is repulsive, so that it is in equilibrium with a third charge (q₃) this must be of the opposite sign and be between the two charge (q)

let's apply Newton's second law to one of the charges, for example the one on the left

         -F₁₂ + F₁₃ = 0

           F₁₂ = F₁₃

          k q₁ q₂ / r₁₂² = k q₁ q₃ / r₁₃²

          q₂ / r₁₂² = q₃ / r₁₃²

          q₃ = q₂ (r₁₃ / r₁₂)²

The problem indicates the charge q₁ = q₂ = 4 q and the distance between them is r₁₂ = L = 9 cm = 0.09 m, we substitute

          q₃ = 4q (r₁₃ / L)²

Let's analyze the situation a bit that the charge 1 and 2 are in equilibrium with a single charge 3 this must be symmetrical between the two charge (the same force), therefore its position on the x axis must be r₁₃ = L/2 and how it is on the y axis = 0

let's substitute

           q₃ = 4q (L / 2L)²

            q₃ = 4q 1/4

            q₃ = q

the charge q3 is ¼ of the charge q

Answer:

Weight of the woman in Newton

   [tex]x = 698.6 \ N[/tex]

 Mass of the woman in slug

  [tex]Mass = 4.86 \ slug[/tex]

 Mass of the woman in kg

   [tex]Mass = 16 \ kg[/tex]

  My weight in Newton

  [tex]W = 784 \ N[/tex]

Explanation:

From the question we are told that

   The weight of the woman in pounds is  [tex]W = 157 \ lb[/tex]

Converting to Newton

    1 N  =  0.22472 lb

    x N  =  157

=>   [tex]x = \frac{157 * 1}{0.22472}[/tex]

=>   [tex]x = 698.6 \ N[/tex]

Obtaining the mass in slug

   [tex]Mass = \frac{W}{g}[/tex]

Here  [tex]g = 32.2 ft/s^2[/tex]

So

     [tex]Mass = \frac{157 }{32.2}[/tex]

       [tex]Mass = 4.86 \ slug[/tex]

Obtaining the mass in kilogram

     [tex]Mass = \frac{W}{g}[/tex]

Here  [tex]g = 9.8 \ m/s[/tex]

So

   [tex]Mass = \frac{157 }{9.8}[/tex]

   [tex]Mass = 16 \ kg[/tex]

Generally weight is mathematically represented as

     [tex]W = m * g[/tex]

Given that my mass is   80 kg   then my weight is  

     [tex]W = 80 *9.8[/tex]

      [tex]W = 784 \ N[/tex]

Answer:

(c) is accelerating upward

Explanation:

When a person stand in an elevator moving upwards, he feels heavier because the elevator's floor presses harder on his feet, and the scale will show a higher reading than when the elevator is at rest.

According to Newton's second law

R = mg + ma

where;

R is the reading of the scale = apparent weight of the person

mg is the normal weight of the person

ma is the upward force acting on the person

Therefore, his apparent weight will be the greatest when the elevator is accelerating upward

(c) is accelerating upward

Answer:

0.98°

Explanation:

First we find the refracting angle of orange in water

Which is given as

စr= sin^-1. ( 1.456 sin60°/1.33)

= 71.2°

Then that of violet in water

စv=sin^-1. ( 1.468sin60°/1.342)

= 72.3°

So angle between boths colours is the difference

71.2- 72.3= 0.98°

Answer:

40 Ω

Explanation:

The following data were obtained from the question:

Potential difference (V) = 20 V

Current (I) = 0.50 A

Resistance (R) =?

From ohm's law:

V = IR

Where:

V is the potential difference.

I is the current.

R is the resistance.

With the above formula, we can obtain the resistance of the resistor as follow:

Potential difference (V) = 20 V

Current (I) = 0.50 A

Resistance (R) =?

V = IR

20 = 0.5 × R

Divide both side by 0.5

R = 20/0.5

R = 40 Ω

Therefore, the resistance of the resistor is 40 Ω

Answer:

Explanation:

An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.

The change in moment is given by the expression below

Δp=m⋅(Δv)=m⋅(vf−vi) .

Given data

mass m= 0.140-kg

initial velocity vi= 120 m/s

final velocity vf= 1 m/s

substituting we have

Δp=m⋅(Δv)=0.14⋅(1−1.2)

Δp=m⋅(Δv)=0.14⋅(-0.2)

Δp= 0.028 kg m/s

The change in momentum was found to be Δp= 0.028 kg m/s

Answer:

Explanation:

according to ohms law we know that

v=IR

given current =2 amps

given resistance =6Ω

so voltage is

v=2*6 =12 V

The voltage across the circuit is equal to 12 V when 2 A current flows through the circuit with a 6 Ω resistor. Therefore, option (B) is correct.

Ohm’s law can be described as the voltage difference across any conductor being directly proportional to the electric current flowing through that conductor while assuming temperature and all the other physical conditions remain constant.

Ohm’s law can apply only if the temperature is kept constant. Ohm’s S.I. unit is rho represented by the symbol Ω, and the current raises the temperature.

The mathematical expression of potential difference given by Ohm's law is:

V = IR

Where V is the voltage and R is the resistance and I the current.

Given, the resistance of the circuit, R = 6 ohm

The current flowing through the circuit, I = 2 A

The potential or voltage difference across the electric circuit will be:

V = I × R

V = 2 × 6

V = 12 Volt

Therefore, the potential difference across the electric circuit is equal to 12 V.

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Answer:

E₁ / E₂ = M / m

Explanation:

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d  / M  

u² = 2 x E₁ e d  / M

Similarly for electrons

u² = 2 x E₂ e d  / m

Hence

2 x E₁ e d  / M =  2 x E₂ e d  / m

E₁ / E₂ = M / m

Answer:

Pure substance are made of only one kind of particles and fixed constant.

Explanation:

Pure substance are combined in fixed ratio into separate by chemical methods into a physical and chemical methods.

Answer:

A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two

Explanation:

Answer:

The  wavelength is   [tex]\lambda = 0.80 \ m[/tex]

The  frequency is  [tex]f = 3.25 \ Hz[/tex]

Explanation:

From the question we are told that

     The distance between two successive crest is 0.80 m

      The  velocity is  v =  2.6 m/s

Generally wavelength is the distance between two successive crest

So  

    [tex]\lambda = 0.80 \ m[/tex]

Now  the frequency is mathematically represented  as

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

        [tex]f = \frac{2.6}{0.80 }[/tex]

       [tex]f = 3.25 \ Hz[/tex]

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

Explanation:

Answer:

Density, d = 5.37 g/cm³

Explanation:

Given that,

Diameter of coin is 19.55 mm

Thickness of the cone is 1.55 mm

Mass of the coin is 2.5 g

We need to find its density. Density is equal to mass upon volume. So,

[tex]d=\dfrac{m}{V}[/tex]

V is volume, V = A d, d is thickness

Radius = d/2 = (1.955/2) cm=0.9775 cm

Thickness, d = 1.55 mm = 0.155 cm

[tex]d=\dfrac{m}{\pi r^2 \times d}\\\\d=\dfrac{2.5}{\pi (0.9775\ cm)^2\times 0.155\ cm}\\\\d=5.37\ g/cm^3[/tex]

So, the density of the coin is 5.37 g/cm³.