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3 years ago

How do organisms use communication to survive?

Physics

1 answer:

just olya [345]3 years ago

3 0

Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.

Explanation:

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An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75

katen-ka-za [31]

Answer:

Part A) 3899 kPa

Part B) 392.33 kJ/kg

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

$u_{1}$ = 214.07 kJ/kg

$\alpha$ $r_{1}$ = 621.2

The relative specific volume at state 2 is obtained from the compression ratio:

$\alpha$ $r_{2}$ = $\frac{\alpha r_{1} }{r}$

=621.2/ 8

= 77.65

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):

$T_{2}$ = 673 K

$u_{2}$ = 491.2 kJ/kg

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:

$P_{2}$ = $P_{1} r\frac{T_{2} }{T_{1} }$

= 95*8*673/300

= 1705 kPa

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:

$deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}$

= 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):

$T_{3}$ = 1539 K

$\alpha r_{3}$ = 6.588

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:

$P_{3} =P_{2} \frac{T_{3} }{T_{2} }$

= 3899 kPa

Part B

The relative specific volume at state 4 is obtained from the compression ratio:

$\alpha r_{4}= r\alpha r_{3}$

= 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:

$T_{4}$ =775 K

$u_{4}$ = 571.74 kJ/kg

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:

$w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg$

Part C

The thermal efficiency is obtained from the work and the heat input:

η= $\frac{w}{q_{in} }$

=0.523

Part D

The mean effective pressure is determined from its standard relation:

MEP= $\frac{w}{\alpha_{1}- \alpha_{2} }$

= $\frac{w}{\alpha_{1}(1- \frac{1}{r} }$

= $\frac{rwP_{1} }{RT_{1} (r-1) }$

=495 kPa

8 0

4 years ago

Halp meeeeeeeeeeeeeeeeee

Greeley [361]

Answer:

1º all

2º 1

3º 2

4º the same

Explanation:

7 0

3 years ago

When Jacob bats a baseball with a net force of 4.719 N, it accelerates 33 m/s2. What is the mass of the baseball?

andreev551 [17]

Answer:

Mass=0.143 kg

Explanation:

From the Newton's second law of motion the force applied to an object is directly proportional to the acceleration produced and the acceleration is in the direction of the force.

F=ma where F is the force, m is the mass of the object and a i the acceleration.

m=F/a

=4.719N/33 m/s²

=0.143 kg

5 0

3 years ago

A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their

Yuki888 [10]

Answer:

The system is still balanced

Explanation:

If we suppose that the boy weights M and the girl m, and are balanced at distances L1 and L2 from the pivot point respectively, thy will be balanced if the resultant torque of all the farces from the pivot pint is cero:

(1) $T=M*L1-m*L2=0$

now they moved to distances (L1)/2 and (L2)/2, the resultant torque will be:

(2) $T=M*(L1)/2-m*(L2)/2$

1/2 can be taken out as a common factor:

(3) $T=(M*L1-m*L2)*1/2$

As the the inside of the parenthesis equals equation (1) that equals zero, hte whole equiation equals zero:

(3) $T=(0)*1/2=0$

So the system is still balanced

6 0

3 years ago

a 55.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 130 kg football players at a height of 0.800 m . the diamet

suter [353]

The diameter of the football player's piston is 0.55 m

Given that the mass of the cheerleader(m) is 55 kg, mass of football player to be hold (M) is 130 kg, height of the players (h) is 1.30 m, radius of the piston corresponding to the diameter (r) is 0.09 m, Diameter of football player's piston (R), P1 is Pressure on the cheerleader's side, P2 is Pressure on the football player's side

Using Pascal's law,

This law states that if there is a change at a point of a body immersed in a fluid then that change will spread thoroughly to each and every point of the body.

The formula of hydraulic system is,

P1= P2

F1/A1 = F2/A2

mg/πr^2 = 4Mg/πR^2

m/r^2=4M/R^2

R^2=4M×r^2/m

By plugging the values, we get.,

R^2=(4×130×0.09^2)/55

R^2=4.21/55=0.076

R=√0.076 = 0.275 m

Hence, diameter of football player's piston is 0.55 m

Learn more about Hydraulic lift here

brainly.com/question/19052774

#SPJ4

8 0

1 year ago