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Ira Lisetskai [31]
3 years ago
9

Answer the Question Correctly and get brainliest and thank you

Physics
1 answer:
PtichkaEL [24]3 years ago
3 0

Answer:

D. are made of cell

Explanation:

first , cell simply means the basic and structural unit of life ......All living organisms have cells

It could be Unicellular(amoeba), colonial form ( volvox) , filamentous form(spirogyra) or multicellular .......

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Which type of energy is a result of motion
Stells [14]

Kinetic energy I think.


6 0
4 years ago
Read 2 more answers
Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the
Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +delta-p

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +0

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

0-0=h_{2} -400x10^{3}  +\frac{40^{2}-250^{2}      }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg

Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}

The exit area is:

m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}

6 0
3 years ago
Which example represents the impact of religion on culture in the United States? A. children receiving counseling at school B. f
dmitriy555 [2]

Answer:

D. Banks and offices closing on sunday.

Explanation:

Church is usually on sunday and people need a day off to do it because service is about 3 hours long. When banks and offices close on this day it shows how religion has affected our lives and our culture.

5 0
3 years ago
g 1. A mass undergoing simple harmonic motion along the x-axis has a period of T = 0.5 s and an amplitude of 25 mm. Its position
kondor19780726 [428]

Answer:

Explanation:

a )

Amplitude A = 14 mm , angular frequency ω = 2π / T

= 2π / .5

ω = 4π rad /s

φ₀ = initial phase

Putting the given values in the equation

14 = 25 cos(ωt + φ₀ )

14/25 = cosφ₀

φ₀ = 56 degree

x(t) = 25cos(4πt + 56° )

b )

maximum velocity = ω A

=  4π  x 25

100 x 3.14 mm /s

= 314 mm /s

At x = 0 ( equilibrium position or middle point , this velocity is achieved. )

maximun acceleration = ω² A

= 16π² x A

= 16 x 3.14² x 25

= 3943.84 mm / s²

3.9 m / s²

It occurs at x = A or at extreme position.

7 0
3 years ago
Harmonics problem. A square wave of frequency f contains harmonics (sine waves) at f, 3f, 5f, 7f, ... . Suppose a system respond
ira [324]

Answer:

B

Explanation:

A square of frequency of consists of the infinite sum of sine waves, whose frequencies are the odd multiples of the main frequency f i.e f, 3f,5f, 7f, ... . Given that the range of frequencies, to which the system responds is 20-40 kHz, for a square wave of frequency 10kHz we need to look for the harmonics whose frequencies are in the systems' respond range, which are the  harmonics of 20, 30 and 40 kHz

4 0
3 years ago
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