Which option gives an object mass in SI units

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

Curious George is whirling a 2.0 kg bunch of bananas on a smooth floor in a circular path having a radius of 0.50 m. What force

Sedaia [141]

Answer:

The force will be $F_{c}=2.47 N$

Explanation:

Let's use the centripetal force equation.

$F_{c}=m\omega ^{2}R$

Where:

m is the mass of the bunch of bananas

ω is the angular speed

R is the radius

Now, 1 rev every 4 seconds or 0.25 rev/sec is the angular speed, but we need to write this speed in rad per second.

$\omega =0.25\frac{rev}{s}=0.25*2\pi \frac{rad}{s}=1.57 \frac{rad}{s}$

FInally, the force will be:

$F_{c}=2.0*1.57^{2}*0.5$

$F_{c}=2.47 N$

I hope it helps you!

4 0

3 years ago

(1) Develop an equation that relates the rms voltage of a sine wave to its peak-to-peak voltage. a. If a sine wave has a peak-to

Aleks04 [339]

Answer:

(A) Equation will be $v=v_msin\omega t=0.75sin(18840t)$

(B) RMS value of voltage will be 0.530 volt

Explanation:

We have given peak to peak voltage of ac wave = 1.5 volt

Peak to peak voltage of ac wave is equal to 2 times of peak voltage

So $2v_{peak}=1.5volt$

$v_{peak}=\frac{1.5}{2}=0.75volt$

Frequency of ac wave is given f = 3 kHz

So angular frequency $\omega =2\pi f$ = 2×3.14×3000 = 18840 rad/sec

So expression of equation will be $v=v_msin\omega t=0.75sin(18840t)$ ( As phase difference is 0 )

Now we have to find the rms value of voltage

So rms voltage will be equal to $v_{rms}=\frac{v_{peak}}{\sqrt{2}}=\frac{0.75}{1.414}=0.530volt$

4 0

3 years ago

What is the speed of a 180 g ball if its kinetic energy is 100 J?

Vaselesa [24]

Answer:

Velocity, V = 33.33 m/s

Explanation:

Given the following data;

Mass = 180grams to kilograms = 180/1000 = 0.18 kg

Kinetic energy = 100J

To find the speed;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Substituting into the equation, we have;

100 = ½*0.18*V²

Cross-multiplying, we have;

200 = 0.18*V²

V² = 200/0.18

V² = 1111.11

Taking the square root of both sides, we have;

Velocity, V = 33.33 m/s

3 0

3 years ago

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge th

statuscvo [17]

Answer:

P /K = 1,997 10⁻³⁶ s⁻¹

Explanation:

For this exercise let's start by finding the radiation emitted from the accelerator

$\frac{dE}{dt}$ = $\frac{q^{2} a^{2} }{6\pi \epsilon_{o} c^{2} }$

the radius of the orbit is the radius of the accelerator a = r = 0.530 m

let's calculate

\frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]

P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W

Now let's reduce the kinetic energy to SI units

K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J

the fraction of energy emitted is

P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹

P /K = 1,997 10⁻³⁶ s⁻¹

3 0

3 years ago