Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds

Answers

Answer 1

Answer:

1.3 m/s

Explanation:

It is given that,

Mass of bird A, [tex]m_A=2.2\ kg[/tex]

Mass of bird B, [tex]m_B=1.7\ kg[/tex]

Initial speed of bird A is 0 as it was at rest

Initial speed of bird B is 3 m/s

We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,

[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]

So, the center of mass for this system is 1.3 m/s.


Related Questions

A car takes 8 s to increase its velocity from 10 m/s to
30 m/s. What is its average acceration?

Answers

Answer:

Average acceleration is 2.5 m/s^2.

Explanation:

Average acceleration can be found by dividing the change in speed by the change in time.

We know that the initial velocity=10m/s, the final velocity=30m/s and time elapsed is t=8s.

a= (Final velocity-Initial Velocty)/t

a=(30-10)/8=20/8=2.5 m/s^2

A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.

Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?

How far does block two travel, d2 in meters, before coming to rest after the collision?

Answers

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:

Friction

Given that,

[tex]m_{1}[/tex] = 18.5 kg

d = 2.3m

To find,

Acceleration of block 1:

∑[tex]F = ma[/tex]

⇒ -m₁gμ = m₁a

⇒ a = -gμ

⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]

∵ a [tex]= -2.16 m/s^2[/tex]

Now,

To determine the velocity of block one prior to the collision:

We know,

The initial velocity of block 1 = 8.25 m/s

⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx

⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]

∵ [tex]v = 7.63 m/s[/tex]

We also know,

[tex]m_{2}[/tex] = 7.25 kg

Now,

The velocity of block 2 post collision:

⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision

Through this,

⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]

∵[tex]v = 19.5 m/s[/tex]

The distance can be found through:

⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]

⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]

∵ Δ[tex]x = 87.8 m[/tex]

Thus, 19.5 m/s and 87.8 m are the correct answers.

Learn more about "Friction" here:

brainly.com/question/13357196

the radius of the earth social

Answers

6,371km is the radius of the earth

Define reflection of sound?​

Answers

Sound travels in waves different waves are different sounds

Explanation:

When sound travels in a given "medium", it would touch the surface of another "medium" and will bounce back in some other direction, this occurrence is called the reflection of sound.

A curve that has a radius of is banked at an angle of . If a -kg car navigates the curve at without skidding, what is the minimum coefficient of static friction between the pavement and the tires

Answers

Answer:

0.65

Explanation:

For whatever reasons, the parameters are not giving. So, I will assume by myself to make the calculations easier. You can substitute whatever it is to it from your question.

Given that

Radius of the road, r = 63 m

Speed of the car, v = 20 m/s

The relationship between a car that is passing through a curve and it's frictional force is said to be

U(s) * g = v²/r

In the formula above,

U(s) = coefficient of static friction

g = acceleration due to gravity, 9.8 m/s²

v = velocity of the car

r = radius of the road

Now when we substitute the earlier stated values, we have

U(s) * 9.8 = 20² / 63

U(s) * 9.8 = 400 / 63

U(s) * 9.8 = 6.35

U(s) = 6.35 / 9.8

U(s) = 0.65

Thus, our coefficient of static friction, based on the stated values is 0.65

Magnets are usually made up of which material

A. plastic

B. iron ore

C. copper

D. gold​

Answers

Answer:

B. iron ore

Explanation:

Hope this helps

plz mark as brainliest!!!!!!

How much would a 15.0 kg object weigh on that planet? Round the answer to the nearest whole number.

Answers

Answer:

168

Explanation:

Answer: a 15 kg object would weigh the most on Neptune

168 N

2. For a rotating rigid body, which of the following statements is NOT correct?
a. All points along a rotating rigid body move with constant linear speed.
b. Points along a rotating body have same angular velocities.
c. Points along a rotating body move through the same angle in equal time intervals.
d. All points have the same angular acceleration.​

Answers

Answer:

                                                dasgfwe

Explanation:

A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance rl (r (A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E

Answers

Answer:

A ) E/8

Explanation:

If the sphere of radius R  carries charge Q,  then the volumetric charge density is:

ρ₁ = [Q/ (4/3)*π*R³]

Therefore the net charge inside r  ( r < R ) is:

q₁ = ρ * (4/3)*π*r³

And E = K * q₁/r                  K = 9,98 *10⁹ [N*m²/C²]

E = K *  ρ * (4/3)*π*r³/r

E = K *  ρ * (4/3)*π*r²

If now the charge is distributed over a sphere of radius 2R

ρ₂ =  [Q/ (4/3)*π*(2R)³]

ρ₂ =  [Q/ (4/3)*π*8*R³]

Then  ρ₂ < ρ₁    in fact     ρ₂ = (1/8)*ρ₁

The electric field depends on the net charge enclosed by a gaussian surface, and the distance between the net charge and the considered point, ( considering the net charge as being at the center of the gaussian surface) In this case, there was no distance change then

E₂ = E₁/8

The right answer is lyrics  A ) E/8

A boat initially moving at 10 m/s accelerates at 2 m/s for 10 s. What is the velocity of the boat after 10 seconds?

Answers

Answer:

30 m/s

Explanation:

v = u + at

given that,

u = 10 m/s (initial speed)a = 2 m/s^2 t = 10sv =?(final speed)

v = 10 + ( 2 × 10)

v = 10 + 20

v = 30 m/s

A 1200 kg car is accelerating at 4.5 m/s2. What is the force on the car?

Answers

Answer:

5400 N

Explanation:

f=ma

f= 1200*4.5

f=5400N

Your TV has a resistance of 10 ohms and a wall voltage of 120 V. How much current and power does it use

Answers

Answer:

Current used is 12 ampere.

Power is 1440 watts.

Explanation:

To find the current used by the TV.

Current (I) = voltage/resistance

Current= 120/10

Current is 12Ampere.

To get power used by the TV,

Power = voltage × current.

= 120× 12

Power = 1440 watts.

am I right? be honest

Answers

Answer:

I chose c because it is the greater slope at point c

What type of research based on approach that used to describe variables rather than to test a predicted relationship between variables?

Answers

Answer:

Correlational research can be used to see if two variables are related and to make predictions based on this relationship.

Why wouldn't carbon dating work to determine the age of the earth?
A) Carbon dating works best on other planets
B) The half life of carbon is too short
C) The age of the earth cannot be determined
D) The half life of carbon is too long.

Answers

Answer:

The half-life of carbon is too short.

Explanation:

The answer is B.

A car starts from rest and accelerates for 7.2 s with an acceleration of 1.4 m/s2. How far does it travel? Answer in units of m.

Answers

Answer:

xn = 36.28 [m]

Explanation:

To solve this problem we must use the following equation of kinematics, which is ideal for a body that moves with constant acceleration.

[tex]x=x_{0}+(v_{o} *t)+(\frac{1}{2} )*a*t^{2}[/tex]

where:

x - xo = displacement of the car [m]

Vo = initial velocity = 0

t = time = 7.2 [s]

a = acceleration = 1.4 [m/s^2]

The initial velocity is zero, as the car begins its movement from rest.

xn = (x - xo), Now replacing

xn = (0*7.2) + 0.5*1.4*(7.2^2)

xn = 36.28 [m]

What is the energy contained in a 1.30 m3 volume near the Earth's surface due to radiant energy from the Sun

Answers

Answer:

The energy contained is 5.856 x 10⁻ J

Explanation:

Average energy density of electromagnetic radiation per unit volume is given by the equation;

[tex]U_{avg} = \frac{1}{2} \epsilon _o E_o[/tex]²

where;

[tex]\epsilon _o[/tex] is permittivity of free space

[tex]E_o[/tex] is maximum electric field strength, this can be calculated from the intensity of sun reaching the Earth's surface.

[tex]E_o = \sqrt{\frac{2I}{\epsilon_o C} }[/tex]

The intensity of sun reaching the Earth is 1350 W/m²

[tex]E_o = \sqrt{\frac{2*1350}{8.885*10^{-12}*3*10^8 } } \\\\E_o = 1008.96 \ V/m\\[/tex]

Average energy density of electromagnetic radiation per unit volume;

[tex]U_{avg} = \frac{1}{2} \epsilon_o E_o^2\\\\U_{avg} = \frac{1}{2} (8.85*10^{-12})(1008.96)^2\\\\U_{avg} = 4.505 *10^{-6} \ J/m^3[/tex]

The energy contained in a 1.30 m³ volume is given by;

E = (4.505 x 10⁻⁶)(1.3)

E = 5.856 x 10⁻ J

Therefore, the energy contained is 5.856 x 10⁻ J



A football team lost 3 yards on their first play. On the second play, 10 yards were gained. On the third play the quarterback got tackled behind the line of scrimmage and lost 6 yards. What was their net gain or loss after the three plays?​

Answers

+1yd

Explanation:

Kfgjjh

Answer:

+1 YD

Explanation:

-3 + 10 - 6 = 1

Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD.

Answers

Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

Given the parameters

m which is the mass of the body

v which is the velocity of the body too

K.E = kinetic energy

The expression for the kinetic energy of a body is given as

KE= 1/2mv²

A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and cresting a 2800 m pass. (a) Neglecting friction, what should the speed of the car be at the top of the second pass? (b) Find the actual speed of the car if the work due to nonconservative forces is – 5 x106 J.

Answers

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

Isaac walks 4 blocks north. Then he turns around and walks 1 block south.
Which of the following correctly describes Isaac's motion?
A. Isaac walked a distance of 5 blocks, and his displacement was 3
blocks north.
B. Isaac walked a distance of 5 blocks, and his displacement was 5
blocks.
C. Isaac walked a distance of 3 blocks, and his displacement was 3
blocks north.
D. Isaac walked a distance of 3 blocks north, and his displacement
was 5 blocks.

Answers

Answer:

Isaac walked a distance of 5 blocks, and his displacement was 3 blocks north.

Explanation:

Distance is what he covered from the beginning, while displacement was what he covered in a specific direction

3. How are force, work, and power related?

Answers

Answer:

Work is the energy needed to apply a force to move an object a particular distance, where force is parallel to the displacement. Power is the rate at which that work is done.

Answer: Work is the energy needed to apply a force to move an object a particular distance, where force is parallel to the displacement. Power is the rate at which that work is done.

Explanation:

What equation relates mechanical energy, thermal energy, and total energy when there is friction present in a system?

Answers

E total = ME + E thermal Explanation:
APEX

g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answers

Answer:

The average force exerted on the superball by the sidewalk is 0.00122 N.

Explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)

final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)

time of motion, t = 1800 s

The average force exerted on the superball by the sidewalk is given by;

[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]

Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion is reversible but not isothermal and that the temperature decreases as the expansion proceeds. (a) Find an expression

Answers

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

= initial volume of gas  =  

= final volume of gas  =  

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

A spring gun is able to launch a 7.0 gram marble to a vertical height of 22 mmeasured from the compressed point of the marble). If the spring iscompressed 8.0 cm from its relaxed state, what will be the spring constant

Answers

Answer: Spring constant = 472N/m

Explanation:

The change in  gravitational potential energy by the spring is given as = mgh

where m= 7.0 g = 7 X 10 -3kg

g= 9.8m/s

h= 22m

Gravitational potential energy= mgh

= 7.0 x  10^-3 X 9.8  x 22 = 1.5092 J

Remember that  change in  gravitational potential energy by the spring =elastic potential energy

Therefore,  Potential energy P. E = 1/2 K x²

where K= CONSTANT

x= 8.0

2 X 1.5092 J / (8.0 X 10^-2)²= 471.625 ROUNDED TO 472 N/m

Can anybody tell me what I'm suppose to do. I click start the numbers comes up to the right ​

Answers

I’ve have any idea. Sorry bro

QUESTION 2

At the end of an investigation, what should be done if the results do NOT support the hypothesis?

O A Repeat your investigation to ensure your results are accurate and then modify your hypothesis if necessary.

OB. Repeat your investigation over and over again until you get the results that will support your original hypothes

O C. Check your measurement tools to ensure they are working.

OD. Change the topic of your investigation to one that will yield results that support a hypothesis.

Answers

Answer:

A. Repeat your investigation to ensure your results are accurate and then modify your hypothesis if necessary.

Explanation:

Having results that do not support the hypothesis is a common occurrence.

Hypotheses always depend on the data and experiment. If at the end of an investigation the results do not support the hypothesis, the investigation should be be repeated to further confirm this discovery.

And if there is still no correlation, then the hypothesis is not a reasonable explanation for the investigation and should be modified or rejected if necessary.

Someone help

Match the definition to the term.

1.

the noun that follows the verb and answers the questions

whom or what

linking verb

2. a word that expresses action

sentence

3. who or what the sentence is about

predicate noun

4.

a noun that indicates to or for whom or what something is

done

predicate adjective

5. follows a linking verb and renames the subject

direct object

subject

6. follows a linking verb and describes a subject

verb that joins a subject and a predicate noun or predicate

7.

adjective

indirect object

8. expresses a complete thought

verb

Answers

Answer:

Explanation:

No 1 is linked to the option provided at no 5.

A direct object is the noun that follows the verb and answers the question.

No 2 is linked to the option provided at no 8.

A verb is a word that expresses action.

No 3 is linked to the option provided at no 5.

The subject is what or whom the sentence is about.

No 5 is linked to the option provided at no 3.

A predicate noun follows a linking verb, and renames the subject.

No 6 is linked to the option provided at no 4.

A predicate adjective follows a linking verb, and describes a subject.

No 7 is linked to the option provided at no 1.

A linking verb joins a subject and a predicate.

No 8 is linked to the option provided at no 2.

A sentence expresses a complete thought....

Answer:

predicate noun: follows a linking verb and describes a subject

verb: a word that expresses action

direct object: a noun that indicates to or for whom or what something is done

sentence: expresses a complete thought

predicate adjective: verb that joins a subject and a predicate noun or predicate adjective

subject: who or what the sentence is about

indirect object: follows a linking verb and renames the subject

linking verb: the noun that follows the verb and answers the questions whom or what

Hope this helps!

Correct me if I'm wrong:)

An increase in temperature the kinetic energy and average speed of the gas particles. As a result, the pressure on the walls of the container . Answer Bank What temperature must a gas, initially at 10 ∘C, be brought to for the pressure to triple?

Answers

Answer:

a

The pressure will increase

b

[tex]T_2 =  576^oC[/tex]

Explanation:

From the ideal gas law we have that

     [tex]PV  =  nRT[/tex]

We see that the temperature varies directly with the pressure so if there is an increase in temperature that pressure will increase

   The initial  temperature is [tex]T_i  =  10^oC = 10 + 273 =  283 \  K [/tex]

The objective of this solution is to obtain the temperature of the gas where the pressure is tripled

Now from the above equation given that nR and V  are constant  we have that

    [tex]\frac{P}{T}  =  constant[/tex]

=>  [tex]\frac{P_1}{T_1}  =\frac{P_2}{T_2}[/tex]

Let assume the initial  pressure is [tex]P_1 =  1 Pa[/tex]

So tripling it will result  to the pressure being [tex]P_2 =  3 Pa[/tex]

So

     [tex]\frac{1}{283}  =\frac{3}{T_2}[/tex]  

=>   [tex]T_2  =  3 *  283[/tex]

=>    [tex]T_2  =  3 *  283[/tex]

=>    [tex]T_2  = 849 \ K [/tex]

Converting back to [tex]^oC[/tex]

   [tex]T_2  =  849 -  273[/tex]

=>  [tex]T_2 =  576^oC[/tex]

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