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3 years ago

Please show work on how to get to the answer. When the diameter of a vessel is increased by 2 times, the flow will :

Physics

1 answer:

Sonja [21]3 years ago

6 0

Answer:

(a) Increase 16 times

Explanation:

According to the Poiseuille flow equation the rate of flow of fluid is directly related to four time square of the radius of the vessel.

$F\alpha r^{4}$

Here, F is the rate of flow, and r is the radius of vessel.

Now according to the question diameter is increased by 2 times then radius is also increased by two times than radius becomes 2r.

So put this value in the flow equation.

$F\alpha (2r)^{4}\\F\alpha 16r^{4}$

Therefore, the rate of flow is increased by 16 times.

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What are the conditions necessary for a terrestrial planet to have a strong magnetic field?.

Stells [14]

Both a molten metallic core and reasonably fast rotation.

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3 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

3 years ago

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$

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3 years ago

You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what

Tatiana [17]

27.9 idkkkk look it up on photomath

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3 years ago

A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

5 0

3 years ago