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emmainna [20.7K]

3 years ago

5

Air enters into the hollow propeller tube at A with a mass flow of 4 kg/s and exits at the ends B and C with a velocity of 400 m

/s, measured relative to the tube. If the tube rotates at 1500 rev/min, determine the frictional torque M on the tube.

1 answer:

mash [69]3 years ago

6 0

Answer:

643N.m

Explanation:

From this question we have:

Mass flow = 4kg/s

Velocity V = 400m/s

Rotation N = 1500rev/min

We get the relative velocity at exit to be:

V2 = V - r2w

400-0.5x [(2*π*1500)/60]

= 400-78.5

= 321.5m/s

Then we have to calculate the frictional torque My

Mt = Mr2 x V2

= 4x0.5x321.5

= 643Nm

From the calculations above, we get the frictional torque M on the tube to be 643Nm.

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A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery

GuDViN [60]

Answer:

at the beginning: $2.3\cdot 10^{-10} F$

when the plates are pulled apart: $1.1\cdot 10^{-10} F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 \frac{A}{d}$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^{-3} m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F$

4 0

2 years ago

A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th

inn [45]

Answer:

$|\vec r|=339.82\ m$

$\theta=-6.67^o$

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

$\vec r=\vec r_1+\vec r_2$

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

$x=rcos\theta$

$y=rsin\theta$

And the vector is expressed as

$\vec z==$

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

$\vec r_1==\ km=\ m$

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

$\vec r_2==\ m$

The total displacement is

$\vec r=\ m+\ m$

$\vec r=\ m$

In (magnitude,angle) form:

$|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m$

$\boxed{|\vec r|=339.82\ m}$

$\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169$

$\boxed{\theta=-6.67^o}$

5 0

3 years ago

Does Ap Physics have anything to do with probablity?

padilas [110]

So I'm a junior. I am currently taking AP Calc BC and AP Physics B.

As of now, I'm not sure if I should take AP Probability and Statistics or Differential Equations/Calc III next year. Also, I'm debating between taking AP Physics C or AP Chemistry.

Which ones do you think would look better on a transcript? I heard that Diffeq/CalcIII is harder than AP ProbStat, but ProbStat is an AP course which will be weighted heavier. Also, should I take Physics C since i've taken Physics B this year already?

5 0

3 years ago

A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070

likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

8 0

3 years ago

Suppose that the net resistance comes from two resistors in series: R=R1+R2. Assume that you measure each resistor independently

rewona [7]

Answer:

± (.021 ) ohm

Explanation:

In the addition of two physical quantities , the uncertainties are simply added .

So , net uncertainty in the value of R will be

± (.007 +.014)

=± (.021 ) ohm

8 0

3 years ago

Read 2 more answers