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2 years ago

9

on takeoff, the propellers on a uav (unmanned aerial vehicle) increase their angular velocity from rest at a rate of ω = (25.0t)

rad/s for 3.0 s. (a) what is the instantaneous angular velocity of the propellers at t = 2.0 s? (b) what is the angular acceleration?

1 answer:

Rudik [331]2 years ago

7 0

Hi there!

a)
We can find the angular velocity at t = 2.0 s by plugging in this value into the equation.

$\omega (t) = 25.0t \\\\\omega (2) = 25.0(2) = \boxed{50.0 \frac{rad}{s}}$

b)

The angular acceleration is the derivative of the angular velocity, so:
$\alpha (t) = \frac{d\omega}{dt} (25.0t) = 25.0$

Thus, the angular acceleration is a <u>constant 25 rad/s².</u>

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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

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Answer:

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As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

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Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

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So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

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4 years ago

A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².

Dmitry_Shevchenko [17]

Answer:

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To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

$P=F/A$

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But first we must convert the units from square millimeters to square meters.

$A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2} } =0.0001[m^{2} ]$

Now replacing:

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Answer:

<h3>The answer is 3 kg</h3>

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