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Nana76 [90]
11 months ago
9

determine the magnitude of the flux in kg/s (no sign, just the magnitude) of air through the surface s. the density of air is 1.

2 kg/m3. the velocity of the air is
Physics
2 answers:
Nonamiya [84]11 months ago
5 0

the velocity of the air is94.25V kg/s

Calculation :

Assuming velocity of air through spherical shell = V m/s

then flux = ρ *π *a^{2}*V kg/s

=1.2 *π *5^{2}*V  = 94.25V kg/s

Rate of change of distance. Rate of change of displacement. The velocity of a moving object can never be negative.

Velocity is the directional velocity of a moving object as an indicator of the observed rate of change of position from a given reference frame.

Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) versus change in time (Δt), given by the formula v = Δs/Δt.

Learn more about Velocity here : brainly.com/question/25522879

#SPJ4

madreJ [45]11 months ago
3 0

The correct answer is 94.25 kg/s

Flux can be stated as the rate of flow of a fluid, radiant energy, or particles across a given area.

The magnetic flux is the magnetic lines present in among two magnets or solenoid is mutual flux.

Velocity is the count of difference in displacement in a given time and it is a vector quantity which has both magnitude and direction.

The mass of a sample of a chemical compound divided by the quantity, or number of moles in the sample, measured in moles, is known as the molar mass of that compound.

Given

density = 1.2 kg/m³

The velocity of the air is 94.25V kg/s

Assuming velocity of air through spherical shell = V m/s

Then flux = ρ *π * a² * V kg/s = 94.25V kg/s

Therefore, the velocity given will be 94.25V kg/s

To know more about velocity, refer: brainly.com/question/28965269

#SPJ4

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V is greater

Explanation:

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
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A child wearing socks is trying to see how far he can slide across his kitchen floor. If
Ugo [173]

Answer:

The coefficient of static friction between the ground and the soles of a runner’s shoes is 0.98. What is the maximum speed in which the runner can accelerate without slipping if they have a mass of 73 kg?

Explanation:

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if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *
SSSSS [86.1K]

Answer:

0.013N

Explanation:

F=\frac{mv^{2} }{r}

m=0.04, v=4.4m/s, r=60

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