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3 years ago

Why do organisms differ in their methods of reproduction?

2 answers:

3 0

Asexual reproduction yields genetically-identical organisms because an individual reproduces without another. In sexual reproduction, the genetic material of two individuals from the same species combines to produce genetically-different offspring; this ensures mixing of the gene pool of the species.

Dmitrij [34]3 years ago

3 0

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Steam flowing through a long, thin walled pipe maintains the pipe wall at a uniform temperature of 500 K. The pipe is covered wi

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Answer: I would love to learn this

Explanation:

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3 years ago

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3 years ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

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3 years ago

A spherical steel container 3 feet in diameter is buried in a land fill. The container is filled with a chemical that keeps the

zmey [24]

Answer:

Q = 378.247 Bt/hr

Explanation:

given data:

diameter of container = 3 m

so r = 1.5 m

T1 = 50°C

T2 = 100°C

depth y = 3 ft

Heat transfer is given as Q

$Q = SK\Delta T$

Where

S = Shape factor for the object

$S = \frac{4\pi r}{1-\frac{r}{2y}}$

$S = \frac{4\pi *1.5}{1-\frac{1.5}{2*3}}$

S = 25.132 ft

$Q = SK\Delta T$

Q = 25.132*0.301 *(100-50)

Q = 378.247 Bt/hr

5 0

3 years ago

For the following pairs of sinusoidal time functions, determine which one leads/lags and by how much. (a) ????1(????)=4sin(6π×10

rjkz [21]

Answer:

The question is incomplete, the complete question is given below

"For the following pairs of sinusoidal time functions, determine which one leads/lags and by how much. (a) V1(t) =4sin(6π×10^4t+60°)V and V(t)2=2cos(6π×10^4t−20°)V. (b) V(t)=10cos(400t−75°) V and I(t)=4sin(400t+30°) A.

Answer

A. V2(t) leads V1(t) by 10°

B. I(t) leads V(t) by 15°

Explanation:

First we express the relationship between sine and cosine of a value.

The expression is giving below Cos (wt) =Sin(wt+90)

Hence for the equations above, we write

a. We can v(t) as

V1(t)=4Sin(6π*10^4+90°-30°)

V1(t)=4Cos(6π*10^4-30°)

Comparing to

V2(t)=4Cos(6π*10^4-20°)

Comparing the angle, we notice that V2(t) leads V1(t) by 10°

b. We can write the current wave form as

I(t)=4sin(400t+90°-60°)

I(t)=4Cos(400t-60°)

If we compare with V(t)=10cos(400t−75°)

I.e 4Cos(400t-60°)=10cos(400t−75°)

We can conclude that I(t) leads V(t) by 15°

4 0

3 years ago