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Gelneren [198K]
2 years ago
10

Which trait shared by dolphins and bats possibly lead to the evolution of echolocation in these two animal groups? the need to m

ove quickly through dark environments the need to be able to send sound waves through the air the need to be able to send sound waves through the water the need to communicate with other species through echolocation.
Physics
1 answer:
Lisa [10]2 years ago
3 0

A trait shared by dolphins and bats that possibly led to the evolution of echolocation in these two animal groups will be the need to move quickly through dark environments.

<h3>What is the evolution of echolocation?</h3>

Our understanding of the evolution of echolocation in bats has shifted as a result of recent molecular phylogenies. These phylogenies imply that bats with advanced echolocation

According to one interpretation of these trees, laryngeal echolocation originated in the ancestor of all living bats. Echolocation may have been lost in Old World fruit bats

The vast adaptive radiation in echolocation call design is substantially controlled by ecology, demonstrating how environmental perceptual problems influence call design.

A trait shared by dolphins and bats that possibly led to the evolution of echolocation in these two animal groups will be the need to move quickly through dark environments.

Hence option A is correct.

To learn more about the evolution of echolocation refer to the link;

brainly.com/question/20789287

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Answer:

Answers can be seen below

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1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

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Vesnalui [34]

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle (\Phi) = 20°

Pinion speed (n_p ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear (N_G) = 50

Teeth on pinion (N_p) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = \frac{N}{P}

=\frac{20}{5}

= 4 in

From the values of Lewis Form Factor Y for (n_p ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

V = \frac{\pi d n_p}{12}

V = \frac{\pi *4*2000}{12}

V = 2094.40 ft/min

For cut or milled profile; the velocity factor (K_v) can be determined as follows:

K_v = \frac{2000+V}{2000}

K_v = \frac{2000+2094.40}{2000}

= 2.0472

However, there is need to get the value of the tangential load(W^t), in order to achieve that, we have the following expression

W^t=\frac{T}{\frac{d}{2} }

W^t = \frac{63025*H}{\frac{n_pd}{2}}

W^t = \frac{63025*30}{2000*\frac{4}{2}}

W^t = 472.69 lbf

Finally, the bending stress is calculated via the formula:

\sigma = \frac{K_vW^tp}{FY}

\sigma = \frac{2.0472*472.69*5}{1*0.321}

\sigma = 15073.07 psi

\sigma = 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

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3 years ago
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