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3 years ago

A scientist is observing a eukarotic cell and a prokaryotic cell. Which structure could she only observe in the eukaryotic cell?

Physics

2 answers:

Ksenya-84 [330]3 years ago

7 0

The answer is nucleus

tangare [24]3 years ago

4 0

Answer:

It's definitely Nucleus

Hope this Helps!

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Which of the following best describes the difference between type a and type b

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Is there supposed to be a picture? Next time try putting a picture

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3 years ago

I’m a freshman, I got a 80% in the first quarter, 85% in the second and 70% in the third. I have a 46% in the fourth and 34 tota

Nadusha1986 [10]

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3 years ago

One light-hour is the distance that light travels in an hour. How far is this, in kilometers? (Recall that the speed of light is

ycow [4]

Answer:

B 1.08 BILLION

Explanation:

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3 years ago

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)

levacccp [35]

In component form, the displacement vectors become

• 350 m [S] ==> (0, -350) m

• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

5 0

3 years ago

(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet

trapecia [35]

Answer:

a ) $\dot m = 351.49 kg/s$

b) $\dot m_{actual} = 1046.15 kg/s$

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature = 1000 K

AT the end of isentropic process (compression) temperature is

$\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}$

$T_2' = 610.181 K$

AT the end of isentropic process (expansion) temperature is

$\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}$

$T_4' = 491.66 K$

isentropic work is given as

$w(compressor) = CP (T_2' -T_1)$

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

$\dot m = \frac{70000}{510.88 - 311.73}$

$\dot m = 351.49 kg/s$

b) actual mass flow rate uis given as

$\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}$

$\dot m_{actual} = 1046.15 kg/s$

6 0

3 years ago