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babymother [125]

2 years ago

13

1. Choose the correct answer 1. Casparian strips are present in the of the root. a) cortex b) pith c) pericycle d) endodermis 2.

The endarch condition is the characteristic feature of a) root c) leaves b) stem d) flower

1 answer:

zloy xaker [14]2 years ago

4 0

Answer:

1 casparian strips are present in the root of endodermis.

2 the endarch condition is the character fature of stem.

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A transverse wave on a string is described by the wave function

svp [43]

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.

Read more on transverse wave here

brainly.com/question/2516098

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5 0

1 year ago

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

2 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

3 0

2 years ago

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five

Keith_Richards [23]

Answer:

<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

<em></em>

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = $I$

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = $5I$ (five times larger)

angular speed = ω/5 (five times smaller)

Recall that the kinetic energy of a spinning body is given as

$KE = \frac{1}{2}Iw^{2}$

therefore,

for the first case, the K.E. is given as

$KE = \frac{1}{2}Iw^{2}$

and for the second case, the K.E. is given as

$KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}$

$KE = \frac{1}{10}Iw^{2}$

<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>

<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

6 0

3 years ago

Which state of matter has the greatest distance between the individual particles?

Maurinko [17]

Your answer is c steam because steam is a gas...

4 0

3 years ago

Read 2 more answers