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forsale [732]
3 years ago
12

The Michelson-Morley experiment was designed to measure Group of answer choices the velocity of the Earth relative to the ether.

the relativistic momentum of the electron. the relativistic mass of the electron. the acceleration of gravity on the Earth's surface. the relativistic energy of the electron.
Physics
1 answer:
NISA [10]3 years ago
4 0

Answer:

The Michelson-Morley was designed to detect the motion of the earth through the ether.

No such relation was found and the speed of light is assumed to be the same in all reference frames.

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If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou
dusya [7]

Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

Frequency = 4.40 MHz

Speed of wave = 1540 m/s

We need to calculate the frequency

Case (I),

Observer is moving away from the source

Using Doppler's effect

f'=\dfrac{v-v'}{v}f

Where, v' = speed of observer

Put the value into the formula

f'=\dfrac{1540-0.32}{1540}\times4.40

f'=4.399\ MHz

Case (II),

Cell is as the source of sound of frequency f' and it moving away from the observer.

Using formula of frequency

f''=\dfrac{v-v_{s}}{v+v_{s}}\times f

f''=\dfrac{1540-0.32}{1540+0.32}\times4.399

f''=4.3971\ MHz

We need to calculate the beat frequency

\Delta f= f'-f''

\Delta f=4.399-4.3971=0.0019\ MHz

Hence, The beat frequency is 0.0019 MHz.

4 0
3 years ago
A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against
valkas [14]

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ            

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies V_1=12.7dm^3=12.7liters

And volume is increased by 3.3dm^3 so V_2=12.7+3.3=16liters

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj

(b) At reversible process work done is given by W=nRTln\frac{V_2}{V_1}

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125

So work done W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J

7 0
3 years ago
The curvature of the helix r​(t)equals(a cosine t )iplus(a sine t )jplusbt k​ (a,bgreater than or equals​0) is kappaequalsStartF
4vir4ik [10]

Answer:

\kappa = \frac{1}{2 b}

Explanation:

The equation for kappa ( κ) is

\kappa = \frac{a}{a^2 + b^2}

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

\kappa (a) = \frac{a}{a^2 + b^2}

Now, the conditions to find a maximum at a_0 are:

\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0

\frac{d^2\kappa(a)}{da^2}  \left | _{a=a_0} < 0

Taking the first derivative:

\frac{d}{da} \kappa = \frac{d}{da}  (\frac{a}{a^2 + b^2})

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da}  (\frac{1}{a^2 + b^2} )

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1)  (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da}  (a^2+b^2)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a  (\frac{1}{(a^2 + b^2)^2} ) (2* a)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2}  -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 -  2 a^2)

\frac{d}{da} \kappa = \frac{b^2 -  a^2}{(a^2 + b^2)^2}

This clearly will be zero when

a^2 = b^2

as both are greater (or equal) than zero, this implies

a=b

The second derivative is

\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 -  a^2}{(a^2 + b^2)^2} )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 -  a^2 ) + (b^2 -  a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2}  )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2  a ) + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

\frac{d^2}{da^2} \kappa = \frac{-2  a}{(a^2 + b^2)^2} + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

We dcan skip solving the equation noting that, if a=b, then

b^2 -  a^2 = 0

at this point, this give us only the first term

\frac{d^2}{da^2} \kappa = \frac{- 2  a}{(a^2 + a^2)^2}

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

\kappa = \frac{b}{b^2 + b^2}

\kappa = \frac{b}{2* b^2}

\kappa = \frac{1}{2 b}

3 0
3 years ago
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0
3 years ago
Alien A lifts a 500-newton child from the floor to a height of 0.40 meters in 2 seconds
vivado [14]
Strong alien you got there good luck bud you never asked a question
4 0
3 years ago
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