An airplane of mass 13300 kg is flying in a straight line at a constant altitude and with a speed of 560.0 km/hr. The force that keeps the airplane in the air is provided entirely by the aerodynamic lift generated by the wings. The direction of this force is perpendicular to the wing surface. Calculate the magnitude of the lift generated by the wings of this airplane.

Answers

Answer 1

Answer:

The magnitude of the lift generated by the wings of the airplane is 130,340 N.

Explanation:

Given;

mass of the airplane, m = 13,300 kg

speed of the airplane, v = 560 km/h = 155.56 m/s

The magnitude of the lift generated by the wings of the airplane is calculated as;

[tex]F_l = mg\\\\where;\\\\F_l \ is \ the \ magnitude \ of \ the \ lift \ generated\\g \ is \ acceleration \ due \ to \ gravity = 9.8 \ m/s^2\\\\F_l = 13,300 \times \ 9.8\\\\F_l = 130,340 \ N[/tex]

Therefore, the magnitude of the lift generated by the wings of the airplane is 130,340 N.


Related Questions

When the disks collide and stick together, their temperature rises. Calculate the increase in internal energy of the disks, assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.)

Answers

Answer:

   ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]    

Explanation:

This is an interesting problem, no data is given, so the result is a general expression.

Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂

we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved

       

initial instant. Just before the crash

           L₀ = L₁ + L₂

with

           L₁ =  I₁ w₁

the moment of inertia of a disc with an axis passing through its center is

           I₁ = ½ m₁ r₁²

we substitute

           I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂

final instant. Right after the crash

         L_f = I w

     

in angular momentum it is a scalar quantity, so it is additive

         I = I₁ + I₂

angular momentum is conserved

         L₀ = L_f

         I₁ w₁ + I₂ w₂ = I w

         w =  [tex]\frac{ I_1 w_1 + I_2 w_2 }{I}[/tex]            (1)

We already have the angular velocities of the system, let's find the kinetic energy of it

initial

          K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²

final

          K_f = K = ½ I w²

the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses

         ΔK = K_f -K₀

         ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²)           (2)

In this chaos we know all the values ​​for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated

Now if all this variation of energy turns into heat

         Q = ΔK

         m_{total} c_e ΔT = ΔK

where the specific heat of the bear discs must be known, suppose they are of the same material

         ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]               (3)

to make a special case, we suppose some data

the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC

we look for the angular velocity

          I₁ = I₂ = I₀

          I = 2 I₀

         

we substitute in 1

           w = [tex]\frac{I_o w_1 + I_o 0 }{2I_o}[/tex] I₀ w₁ + I₀ 0 / 2Io

           w = w₁ /2

we look for the variation of the kinetic energy with 2

       

             ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)

             ΔK = ¼ I₀ w₁² -½ I₀ w₁²

             ΔK = - ¼ I₀ w₁²

the negative sign indicates that the kinetic energy decreases

We look for the change in Temperature with the expression 3

              ΔT = [tex]\frac{ \Delta K}{(m_1 +m_2) c_e}[/tex]ΔK / (m1 + m2) ce

              ΔT = [tex]\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}[/tex]

              ΔT =  [tex]\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}[/tex]

              ΔT = [tex]\frac{1}{16} r_1^2 w_1^2 / c_e[/tex]

in this expression all the terms are contained

The increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].

What is internal energy?

The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.

The given data in the problem is;

[tex]\rm \omega_1[/tex]  is the angular velocity of  disk 1

[tex]\rm \omega_2[/tex] is the  angular velocity of disk 2

r₁ is the radius of  disk 1

r₂ is the radius of  disk 2

m₁ is the mass of disk 1

m₂ is the mass of disk 2

Momentum before the collision;

[tex]\rm L_1 = I_1 \omega_1[/tex]

The moment of inertia of disc 1

[tex]\rm i_1 = \frac{1}{2} m_1r_1^2[/tex]

The momentum gets conserved;

[tex]\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}[/tex]

The change in the kinetic energy is;

[tex]\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )[/tex]

The change in the energy gets converted into heat;

[tex]\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k[/tex]

The change in the temperature is

[tex]\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}[/tex]

The internal energy change is found by;

[tex]\rm \triangle E = mc_v dt[/tex]

[tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex]

Hence the increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].

To learn more about the internal energy refer to the link;

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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 734.1 rad/s 734.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds0.569 seconds , what is the magnitude of the average angular acceleration of the disk

Answers

Answer:

[tex]1290.16\ \text{rad/s}^2[/tex]

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = 734.1 rad/s

[tex]\omega_f[/tex] = Final angular velocity = 0

t = Time = 0.569 seconds

[tex]\alpha[/tex] = Angular acceleration

From the kinematic equations of rotational motion we have

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-734.1}{0.569}\\\Rightarrow \alpha=-1290.16\ \text{rad/s}^2[/tex]

The magnitude of the average angular acceleration of the disk is [tex]1290.16\ \text{rad/s}^2[/tex].

A 744 N force is applied to an object to reach an acceleration of 24 m/s2. What is the objects mass?

Answers

31kg

Explanation:

F = ma

m = F/a

m = 744N/24m/s^2

m = 31kg

(*Newton's Second Law*)

A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebble of mass m2 from a height of 4h, and it hits the floor with the same kinetic energy KE. How do the masses of the pebbles compare

Answers

Answer:

Hello,

QUESTION)

✔ We have: KE = PE (potential energy)

PE = m x g x h

The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2  

PE1 = PE2 ⇔ PE1/PE2 = 1

[tex]\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\ \frac{m_1}{m_2\times 4} = 1 \\ \\ \frac{m_1}{m_2} = 4[/tex]

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

 

Fill in the graph for 50 points

Answers

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Explanation:

The way you change the speed of a wave is to:

a
Change it's medium
b
Change it's energy
c
Transfer it to a new position
d
Apply a force

Answers

Answer:

transfer it to a new position

Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart

Answers

Answer:

in this case the weight of the vehicle does not change , consequently the friction force should not change

Explanation:

The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation

          fr = μ N

          W-N= 0

          N = W

as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change

True or false? A system must contain more than one object.

Answers

Answer:

true

Explanation:

normally -No system has ever performed well with one object.

A system must contain more than one object is a true statement.

What is system?

A system is a group of interacting or interrelated objects that act according to a set of rules to form a unified whole.

Normally, no system has ever performed well with one object.

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A beam of protons is directed in a straight line along the z direction through a region of space in which there are crossed electric and magnetic fields The electric field is 550 V m in the y direction and the protons move at a constant speed of 105 m s 1 What must be the magnitude of the magnetic field such that the beam of protons continues along its straight line trajectory Express your answer using two significant figures

Answers

Answer:

B = 5.23 T

Explanation:

Given that,

Electric field, E = 550 V/m

The speed of the proton, v = 105 m/s

We need to find the magnitude of the magnetic field such that the beam of protons continues along its straight-line trajectory.

To move in a straight line, the magnitude of the electric force from the field and the magnetic field must be equal i.e.

[tex]qE=qvB\\\\B=\dfrac{E}{v}\\\\B=\dfrac{550}{105}\\\\B=5.23\ T[/tex]

So, the magnitude of the magnetic field is equal to 5.23 T.


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pleaseeeeeeee
Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle

Answers

Answer:

[tex]\tau=0.03\ N-m[/tex]

Explanation:

Given that,

Force acting, F = 6N

The radius of the path, [tex]r=10^{-2}\ m[/tex]

Angle, [tex]\theta=30^{\circ}[/tex]

We need to find the amount of torque acting on the object. The formula for torque is given by :

[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]

So, the required torque is equal to 0.03 N-m.

which two options describes behaviors of particles that are related to the chemical properties of the materials

a- forming hydrogen bonds between them
b- reacting quickly with water
c- having a high mass
d- forming bonds with other atoms

Answers

The answer is D I’m pretty sure

Answer:

The two correct answers are B.) reacting quickly with water, and D.) forming bonds with other atoms.

Explanation:

I took the quiz on a.pex and these were correct.

How much work is done in pushing an object 7.0 m across a floor with a force of 50 N and then
pushing it back to its original position? How much power is used if this work is done in 20 sec?

Answers

Answer:

35/2 J/s

Explanation:

Just use the 2 formulas

Work done = Force * distance moved

Power = Work done/time

WD = 7 * 50  = 350

Power = 350 / 20

= 35/2 J/s

30. Easy Guided Online Tutorial One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.

Answers

Answer:

[tex]18.18\ \text{m/s}[/tex]

[tex]6.82\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of large object = 8 kg

[tex]m_2[/tex] = Mass of smaller object = 3 kg

When large mass is moving

[tex]u_1[/tex] = 25 m/s

[tex]u_2[/tex] = 0

For completely inelastic collision we have the relation

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 25+3\times 0}{8+3}\\\Rightarrow v=18.18\ \text{m/s}[/tex]

Speed of the combined mass when the larger object is moving is [tex]18.18\ \text{m/s}[/tex]

When smaller mass is moving

[tex]u_1[/tex] = 0

[tex]u_2[/tex] = 25 m/s

[tex]v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 0+3\times 25}{8+3}\\\Rightarrow v=6.82\ \text{m/s}[/tex]

Speed of the combined mass when the smaller object is moving is [tex]6.82\ \text{m/s}[/tex]

Tom has a mass of 50,000 g and runs up a flight of stairs 4 m high in 12.5 seconds.
Calculate Tom’s power. (g = 10 m/s2)

Answers

Answer:

160 watts.

Explanation:

Remark

Power = Work / Time

Work = F * d

Note: Since he is running up stairs he is doing work against gravity.

Givens

m = 50000 g   kg / 1000 grmsm = 50000 / 1000 = 50 kgh = 4 mtime = 12.5 secondsg = 10 m/s^2

Formula

P = W * d/tW = m*g *d / t

Solution

P = 50kg * 10 m/s^2 * 4 m / 12.5 P = 160 watts.

Betty hits a baseball.
At which point along the baseball's path does it have the most gravitational potential energy?

Answers

Answer:

point 2

Explanation:

ASAP 20 POINTS!!
The air also contained a small amount of argon
As the temperature of the air decreased from 20C to -190 C the argon changed
Explain the changes in arrangement and movement of the particles of the argon as the temperature of the air decreased

Answers

Answer:

See explanation

Explanation:

Let us recall that temperature is a measure of the average kinetic energy of the molecules of a body.The higher the temperature, the higher the kinetic energy of the molecules of the body.

As temperature decreases, the kinetic energy of the molecules of a substance also decreases rapidly and the magnitude of intermolecular interaction between molecules of the substance increases.

Hence, as argon gas is cooled from 20°C to -190°C the kinetic energy of the gas molecules decreases an the magnitude of intermolecular interaction increases hence the gas changes into liquid and subsequently changes into a solid at -190°C.

Badll
Which of the following is an example of
the Law of Inertia?
A. Sitting in a chair and breaking it
B. Throwing a ball in outer space and it goes on forever
unless acted upon by another force
C. Eating a salad to bring chemical energy into the body
D. Driving a car on a track

Answers

Answer: B

Explanation: an object keeps moving unless it’s acted or interrupted by another force
Badll
Which of the following is an example of
the Law of Inertia?
A. Sitting in a chair and breaking it
B. Throwing a ball in outer space and it goes on forever
unless acted upon by another force
C. Eating a salad to bring chemical energy into the body
D. Driving a car on a track.
Answer is B

2. An object is dropped from rest. Calculate its velocity after 2.5s if it is dropped:
a.On Earth, where the acceleration due to gravity is 9.8m/s?
b. On Mars, where the acceleration due to gravity is 3.8m/s?

Answers

Answer:

a=24.5 b=9.5

Explanation:

a forward horizontal force of 50 N is applied to crate a second horizontal force of 180 N is applied to crate in the opposite direction determine the magnitude and direction of the resultant force acting on the crate

Answers

Answer:

130n on the 2nd horizontal

Explanation:

Which of the following would produce the most power?
b
ОООО
A mass of 10 kilograms lifted 10 meters in 10 seconds
A mass of 5 kilograms lifted 10 meters in 5 seconds
A mass of 10 kilograms lifted 10 meters in 5 seconds
A mass of 5 kilograms lifted 5 meters in 10 seconds
d

Answers

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

[tex] Power = \frac {Energy}{time} [/tex]

But Energy = mgh

Substituting into the equation, we have

[tex] Power = \frac {mgh}{time} [/tex]

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

[tex] Power = \frac {10*9.8*10}{5} = 490 Watts [/tex]

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

ocean currents are always cold true or false

Answers

false, since ocean currents act as conveyer belts of warm and cold water.

If you wrap 150 coils of heavy wire around a big iron nail and attach the ends of the wire to a 6.0v battery, you have a A) radio B) electromagnet C) galvanometer D) ammeter​

Answers

Answer:

B

Explanation:

Because of the voltage attached to the iron nail

A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds

Answers

Answer:

[tex]a=2.5\ m/s^2[/tex]

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]

So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].

50 points help please

Answers

Answer:

C?

Explanation:

Yep,It's C all right.

Answer:

Yep,It's C all right.

Explanation:

Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle.

Answers

Answer:

The amount of torque is 0.03 N.m.

Explanation:

To find the amount of torque we need to use the following equation:

[tex] \tau = \vec {r} \times \vec{F} = rFsin(\theta) [/tex]   (1)

Where:

r: is the radius = 1x10⁻² m

F: is the force = 6 N

θ: is the angle = 30°

By entering the above values into equation (1) we have:

[tex]\tau = 1 \cdot 10^{-2} m*6 N*sin(30) = 0.03 N.m[/tex]  

Therefore, the amount of torque is 0.03 N.m.

I hope it helps you!

BRAINLEST FOR CORRECT ANSWER PLEASE

Which has more momentum: a 3 kg sledgehammer swung at 1.5 m/s OR a 4 kg sledgehammer swung at 0.9 m/s? SHOW YOUR WORK

Answers

Answer:

Sledgehammer A has more momentum

Explanation:

Given:

Mass of Sledgehammer A = 3 Kg

Swing speed = 1.5 m/s

Mass of Sledgehammer B = 4 Kg

Swing speed = 0.9 m/s

Find:

More momentum

Computation:

Momentum = mv

Momentum sledgehammer A = 3 x 1.5

Momentum sledgehammer A = 4.5 kg⋅m/s

Momentum sledgehammer B = 4 x 0.9

Momentum sledgehammer B = 3.6 kg⋅m/s

Sledgehammer A has more momentum

The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?

Answers

Answer:

2π/[28 x (10^-3)]

Explanation:

Angular speed : ω=2π/T

T = 28ms = 28 x (10^-3) s

Angular speed = 2π/[28 x (10^-3)]

Match the descriptions with the graphs !

Answers

Answer:

Graph 1 matches with B, 2 with A, and 3 with C.

Explanation:

Graph 2 shows a car whose distance part of the graph is not going up or down, while the time going up. That means that the car is stopped. Graph 1 shows a straight line, meaning that the car is traveling at a constant speed. Graph 3 is a curved line, meaning the speed of the car is changing somehow, and since the line is becoming more horizontal, the car is getting slower.

At the end of the previous experiment, aclumsy scientist drops the coil, while still in the magnetic field, and still oriented with its plane perpendicular to the magnetic field, denting it and changing its shape to a semi-circle. The new shape has the same perimeter, but a different area, and it takes 0.036s to deform. What isthe average induced EMF during this mishap

Answers

Answer:

hello your question has some missing parts below are the missing parts

A Circular, 10-turn coil has a radius of 10.7 cm and is oriented with its plane perpendicular to a 0.2-T magnetic field.

answer : 1 volt

Explanation:

Determine the Average induced EMF during this mishap

A' = A/2  ( for a semi circle )

where A = [tex]\frac{\pi r^2}{2}[/tex]

To determine the Average induced EMF apply the relation below

| E | =  η * [tex]\frac{\beta A}{T}[/tex]   ----- ( 1 )

Replace A in equation 1 with  A = [tex]\frac{\pi r^2}{2}[/tex]

hence equation becomes : | E | =  η * βπr^2 / 2T'

where : T' = 0.0365 ,  β = 0.2 , η = 10 , r = 0.107

∴| E |  = 0.999 ≈ 1volts

A consumer uses 3098 kWh in 29 days. The utility company charges AED 0.077592 per kWh for the electricity plus AED 0.029998 per kWh for the distribution of the electricity . What is the consumer's electric bill for the 29 days?

Answers

for the electricity, multiply 0.077592 by 3098 and you get 240.380016.
for the distribution, multiply 0.029998 by 3098 and you get 92.933804.

you add those two numbers (240.380016 + 92.933804) for the total in 29 days and you get 297.31382 AED

The consumer electric bill for 29 days is AED 331.31

Given that, The utility company charges AED 0.077592 per kWh for the electricity plus AED 0.029998 per kWh for the distribution of the electricity

Since, A consumer uses 3098 kWh in 29 days.

The  utility company charges for electricity is,

                                  [tex]=0.077592*3098=240.38[/tex]

The  utility company charges for distribution of the electricity  is,

                                  [tex]=0.029998*3098=92.93[/tex]

So that, The consumer electric bill is,

                         [tex]=240.38+92.93=333.31[/tex]

Hence, the consumer electric bill for 29 days is AED 331.31

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pls help ill give brainliest if you give a correct answer! Find two integers whose sum is 17 and product is 72pleaseee help Thomas scored 257 points on the first three tests in class. What does he need to score on the next test to have a test average of 85? A. 83 B. 79 C. 87 D. 91Please help me Which of the following is an example of an electromagnetic wave ? 2. Determine the slope of the line that passes throughthe points (-1, -3) and (0, 6). Subtract: (-5x + 8) - (3x + 5) jumbled letters omnoolpiiicst cmoptinoeti Which of the following take place in themitochondria? Select all that apply.glycolysiscitric acid cycleelectron transport chainoxidative phosphorylation find the lateral area of this cone Each state gets the same number of electors as they have ___. As the temperature of a gas is increased, the average kineticenergy of its molecules what is the temperature halfway between 20 degrees and -10 degrees? Which of the following is not a role for protein in the body. acool the body bkeep skin, hair, and nails healthy cbuild muscle dtransport oxygen Joey went to the county fair last weekend. When he got there, he had to walk 1/2 of a mile from the car to the entrance. Then he walked 11/16 of a mile to the carnival rides and 1/2 of a mile from the carnival rides to the farm animal show. Next, he walked 3/4 of a mile back to the car. How many miles did Joey walk in all? If n+1 integers are chosen from the set1,2,3,,2n,where n is a positive integer, must at least one of them be even? Why? 1) Veja o calendrio de plantio e colheita do Arroz na regio Centro-Oeste do Brasil . * Write advantage and disadvantage of spiritual friend? Tambe Electric entered into a written agreement with Home Depot to provide copper wire to Tambe at a price set forth in the writing, and allowed the contractor the option of paying for the wire over a period of time. Tambe later tried to purchase such wire on a payment plan but Home Depot refused. As Home Depot did not fulfill this written agreement, Tambe sued for $68,000, the additional cost it had to subsequently pay to obtain copper wire for its work. Home Depot defended that it had made an oral condition precedent requiring payment in full by Tambe at the time it accepted the price quote in the written agreement. The result is that:_________ Unit 2 Mid-Unit Assessment Please help me with questions 43, 44, and 45