Which wave has the smallest amplitude?

I will give brainliest to the right answer.
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Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

Given the following data:

Note: μs represents microseconds.

Conversion:

1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds

645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

[tex]Frequency = \frac{1}{Period}[/tex]

Substituting the value into the formula, we have;

[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

Conversion:

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = [tex]0.001[/tex] × [tex]1550.39[/tex]

X = 155039 kHz

To 3 significant figures;

Frequency = 155 kHz

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

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Answer:

the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Explanation:

Given the data in the question;

Time period = 6 months = 1.577 × 10⁷ s

orbital speed v = 64000 m/s

since its a circular orbit,

v = 2πr / T

we solve for r

r = vT/ 2π

r = ( 64000 × 1.577 × 10⁷ ) / 2π

r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU

Now, from Kepler's law

T² = r³ / ( m₁ + m₂ )

T = 6 months = 0.5 years

we substitute

(0.5)² = (1.0737)³ / ( m₁ + m₂ )

0.25 = 1.2378 / ( m₁ + m₂ )

( m₁ + m₂ ) = 1.2378 / 0.25

( m₁ + m₂ ) = 4.9512

m₁ = m₂  = 4.9512 / 2 = 2.4756 solar mass

we know that solar mass = 1.989 × 10³⁰ kg

so

m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg

m₁ = m₂ = 4.92 × 10³⁰ kg

Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

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:

[tex]v=(6ti+6k)\ m/s[/tex]

Explanation:

Given that,

The position of a particle is given by :

[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]

Let us assume we need to find its velocity.

We know that,

[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]

So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]

Using the formula

Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

Answer: it would be daddy

Explanation:

Because I’m daddy

Answer:

The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.

Explanation:

Consider newtonian mechanics here.

Dynamic equation is

The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.

we'll consider newtonian mechanics here.

so the dynamic equations is S = ut + 0.5at^2

we know u=0.8

S=1.2+0.9

t=1.2/1.5

find a.

Answer:

the body is subjected to a continuous rotation and the body is not in rotational equilibrium

Explanation:

For an object to have a static equilibrium, it must meet two relationships

             ∑ F = 0

             ∑ τ =0              

force acting on a body fulfills the relation of

         sum F = F - F = 0

when two forces do not move from position.

To find the torque we assume that the counterclockwise rotations are positive

        Σ τ = - F r - F r

        Στ = -2 Fr <> 0

consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium

Answer:

Gap left = Change in length on heating

Gap=Initial length×Coefficient of linear expansion×change in temperature

Gap=10×0.000012×15m

⟹Gap=0.0018 m

this is an example u have to put your equation in it

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

substituting the values of the variables into the equation, we have

k = mg/x  

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

Answer:

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium

[tex]d_0[/tex] is the object distance

[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium

[tex]d_i[/tex] is the image distance

R is the radius of curvature

Now, let [tex]d_0[/tex] = ∞, such that;

[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

we make [tex]d_i[/tex] subject of the formula

[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )

[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )

given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00

so we substitute

[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

[tex]d_i[/tex] = 9.165 / 0.41

[tex]d_i[/tex] = 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm

Answer:

You should multiply 60 kg*9.8 and answer will come.

Hope this will help you.

Answer:

yes she is right you should multiple 60*9.8

have a great day God bless you

[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Answer:

B. Community

Took science classes for 6 years now

Answer:

Accuracy is how much the consequence of an estimation adjusts to the right worth or a norm' and basically alludes to how close an estimation is to its concurred esteem

《OAmalaOHopeO》

Answer:

In a set of measurements, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.

Explanation:

_Hope it helps you_

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

Explanation:

Work cannot be increased by using a machine of some kind.

Work cannot be increased by using a machine of some kind.

A machine is any device in which the effort applied at one end overcomes a load at the other end.

Machines are generally used to perform different tasks faster.

However, a simple machine can not be used to increase the amount of work done at any time.

Force, speed and torque can all be increased using machines.

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Explanation:

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant

Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

Answer:

check photo for solve

Explanation:

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]

Speed of rocket 1 with respect to rocket 2 :

[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]

[tex]$u' = \frac{0.7 c}{1.12}$[/tex]

[tex]u'=0.625 c[/tex]

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

Answer:

Q at the center of the distribution.

Explanation:

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

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The high frequency radio waves used for telecommunications links travel by line of sight and so are obstructed by the curve of the Earth. The purpose of communications satellites is to relay the signal around the curve of the Earth allowing communication between widely separated geographical points.

Explanation:

hope it helps!!

Answer:

in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .

Explanation: