20. For each improvement in glider design, engineers follow
O A. the written instructions that are provided in the hang glider build kit.
O B. an iterative process of testing, modifying, retesting, and modifying again.
O C. a complicated process of checks and balances while obtaining financing.
O D. a mathematical process, rejecting designs that don't follow blueprint dimensions.
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Related Questions
Question 15 of 25
What is the period of a wave that has a frequency of 30 Hz?
Answers
Answer:
0.033 seconds
Explanation:
Period = 1/30 = 0.033 seconds
Answer:
The answer is 0.03 s
Explanation:
A.P.E.X.
Help plz I’ll mark brainliest
Answers
Answer:
The second option- a substance that a wave can travel through.
Explanation:
Hope This Helps!!
(brainliest please)
Question 2 of 25
Which three statements are true about the wave shown?
A. The wave is a longitudinal wave.
B. The wave could be an electromagnetic wave.
C. The wave could be a sound wave.
D. The wave is a mechanical wave.
Answers
Answer:
ACD
Explanation:
according to it's description above
The wave is a mechanical wave and can also be sound wave also the wave is a longitudinal wave. The correct option is A, C, and D.
What is mechanical wave?A mechanical wave is one whose energy cannot be transmitted via a vacuum. To transfer their energy from one place to another, mechanical waves need a medium.
A mechanical wave is something like a sound wave. A vacuum can not be traversed by sound waves.
Waves like mechanical waves require a medium to travel through. Non-mechanical waves are those that can travel through any medium.
Mechanical waves include, but are not limited to, sound waves, water waves, and seismic waves.
The three claims are accurate as follows: The wave is a longitudinal wave and can also be a mechanical wave or a sound wave.
Thus, the correct option is A, C, and D.
For more details regarding mechanical waves, visit:
https://brainly.com/question/24459019
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Your question seems incomplete, the missing image is:
Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron (3_m). It has a positive charge that is three times the magnitude (3_(qe)) of the charge on an electron. What is the ratio of the speed v_c that the cyberon would have when it reaches the upper plate after being released from rest at position h_0 to the speed ve that the electron would have?
Answers
Answer:
The answer is "The last choice".
Explanation:
Please find the complete question in the attachment.
In an external electric field, its electrical energy at positive charge becomes directed to just the electrical domain. Therefore it will speed towards its base plate whenever cyber one is released to rest at h0. It was never going to reach the top plate. Thus, the last choice corrects because in this the cyber-on never reaches its upper stage.
a man is trying to pull a box a distance of 3 m with a force of 20 N that makes a 35º with the horizontal.
Answers
Answer:
34.4Joules
Explanation:
Complete question
a man is trying to pull a box a distance of 3 m with a force of 20 N that makes a 35º with the horizontal. Find the workdone
Work done = Fdsin theta
Force F = 20N
distance d = 3m
theta = 35 degrees
Substitute
Workdone = 20(3)sin 35
Workdone = 60sin35
Workdone = 34.4Joules
Hence the workdone by the man is 34.4Joules
How can you drop two eggs the fewest amount of times, without them breaking?
Answers
I hope this helps.
Answer:
get 2 jugs of water put an egg in each one drop the jugs with parachutes on them in long grass on a sunny non windy day
Explanation:
egg+ground=broken
egg-ground= egg+air
egg+air=unbroken
egg+water= egg+wet
egg+water= unbroken
egg+egg= 2 egg
egg+egg+air= egg+egg+unbroken+unbroken
egg+egg+unbroken+unbroken=(egg+unbroken)2
longgrass+egg= 40%unbroken+60broken+egg
longgrass+egg+egg=20%unbroken+80%broken+2egg
ground+water=mud
mud+egg=unbroken+egg+muddy
air+water=raining
egg+raining+air=wet+egg+slip+50%broken+50%unbroken
ask if need more proof
g Suppose that you seal an ordinary 60W lightbulb and a suitable battery inside a transparent enclosure and suspend the system from a very sensitive balance. (a) Compute the change in the mass of the system if the lamp is on continuously for one year at full power. (b) What difference, if any, would it make if the inner surface of the container were a perfect reflector
Answers
Answer:
kekemeeimdeiddnekem
Explanation:
mdjdjdiddmjd jjeneeiej
Blythe and Geoff compete in a 1.00-km race. Blythe's strategy is to run the first 600 m of the race at a constant speed of 4.10 m/s, and then accelerate to her maximum speed of 7.30 m/s, which takes her 1.00 min, and then finish the race at that speed. Geoff decides to accelerate to his maximum speed of 8.30 m/s at the start of the race and to maintain that speed throughout the rest of the race. It takes Geoff 3.00 min to reach his maximum speed. Assume all accelerations are constant.
Required:
a. Calculate the time of Blythe's run.
b. Calculate the time of Geoff s run.
Answers
Answer:
Explanation:
1.00 km = 1000 m .
Blythe's run : -------
Time taken to run 600 m speed
= distance / speed
T₁= 600 / 4.10 = 146.34 s
Next time T₂ = 1 min = 60 s
acceleration of Blythe
a = (7.30 - 4.10) / 60 = .053 m /s²
displacement during acceleration
= ut + 1/2 at²
= 4.10 x 60 + .5 x .053 x 60²
= 246 + 95.4
= 341.4 m
Rest of the distance to be covered = 1000 - ( 600 + 341.4 )
= 58.9 m
Time taken to cover this distance
T₃= 58.9 / 7.3 = 8.06 s
Total time = T₁ + T₂ + T₃ = 214.4 s
Geoff s run : ---------
initial acceleration during first 3 min
= (8.3 - 0 ) / (3 x 60 )
= .046 m /s²
displacement
s = ut + 1/2 a t²
= 0 + .5 x .046 x ( 3 x 60 )²
= 745.2 m
Rest of the distance of race
= 1000 - 745.2 = 254.8 m
This distance is covered at speed of 8.3 m/s
time taken to cover this distance
T₂ = 254.8 / 8.3
= 30.7 s
Total time taken to complete the race
= 180 + 30.7
= 210.7 s .
The combination of an applied force and a friction force produces a constant torque of 36.0 N⋅m on a wheel rotating on a fixed axis. While the force acts for 6.00 s, the angular velocity of the wheel increases from 0 to 10.0 rad/s. The force is removed and the wheel comes to rest in 60.0 s
a. Find the moment of inertia of the wheel.
b. Find the magnitude of the torque due to friction.
c. Find the total number of rotations during the 66.0 s.
d. Find the Kinetic energy of the wheel at 6.00 s when the force is removed.
Answers
Answer:
D. Find the kinetic energy of the wheel at 6.00 z when the force is removed.
energy in a stretched spring
Answers
Answer:
Elastic potential energy is the potential energy stored by stretching or compressing an elastic object by an external force such as the stretching of a spring. It is equal to the work done to stretch the spring which depends on the spring constant k and the distance stretched.
Hope this helps :)
-ilovejiminssi♡
An object's mass has a greater influence on its kinetic energy than does its velocity. True or False?
Answers
Answer: I think false
Explanation:The velocity at which an object is sent moving and the mass of the object both play a hand in the level of kinetic energy that object produces. Mass and kinetic energy have a positive relationship, which means that as mass increases, kinetic energy increases, if all other factors are held constant.
Potential energy is energy due to motion.
True or False?
Answers
Answer:
true
Explanation:
Answer:
true
Explanation:
please give brainlest need 1
Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown to the right. The mass of the left block m1 = 1.4 kg and the mass of the right block m2 = 4.9 kg. The angle between the applied force and the horizontal is θ = 54°. The coefficient of kinetic friction between the blocks and the surface is μ = 0.38. Each block has an acceleration of a = 3.6 m/s2 to the right.
Answers
Answer:
Explanation:The Mass Of The Left Block M1 = 1.3 Kg And The Mass Of The Right Block M2 = 3.1 Kg. The Angle Between The String And The Horizontal Is ... (10%) Problem 8: Two blocks connected by a string are pulled across a horizontal surface by a ... m m, 50% Part (a) Write an equation for the magnitude of the force exerted by the ...
What are two things that happen to the sugars that are made by the plant during photosynthesis?
I
Answers
Answer:
The sugars produced by photosynthesis can be stored, transported throughout the tree, and converted into energy which is used to power all cellular processes. Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.
Explanation:
I think this is correct lol.
As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on a tall ring that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration and the normal force on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride
Answers
Answer:
Answer is explained in the explanation section below.
Explanation:
In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.
a) You pass through the highest point:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.
And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.
b) You pass through the lowest point of the ride:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.
And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.
a 2,400 kg car drives north towad a 60kg shopping cartthat has a velocity of zero the two objects collide giving the car a final velocity 4.33m/s north and the shopping cart 8.88m/s north what is the in itial velocity of the car
Answers
Answer:
4.552m/s
Explanation:
[tex]V=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1} } =\frac{2400*4.33+60*8.88}{2400}=4.552m/s[/tex]
What is the correct coefficient for 2H2 + O2 →2H2O
Answers
Explanation:
2forH2,1for02,and2forH20
The new springs will be identical to the original springs, except the force constant will be 5655.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height
Answers
Answer:
Explanation:
For original spring , compression in spring due to a load of 1355 kg is
x = 12 - 8.55 = 3.45 cm = .0345 m
spring constant = W / x
= 1355 x 9.8 / .0345
= 384898.55 N /m
Spring constant of new spring
k = 384898.55 - 5655 = 379243.55 N /m
New compression for new spring
= W / k
= 1355 x 9.8 / 379243.55
= .035 m
= 3.50 cm
Difference of compression = 3.50 - 3.45
= .05 cm .
In later case , car will be more lowered by .05 cm .
A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 46.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answers
Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
[tex] -F_{f} + F = 0 [/tex]
[tex] -F_{f} + ma = 0 [/tex]
[tex] \mu mg = ma [/tex]
[tex] \mu = \frac{a}{g} [/tex]
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} [/tex]
Where:
d: is the distance traveled = 46.1 m
[tex] v_{f}[/tex]: is the final speed of the truck = 0 (it stops)
[tex]v_{0}[/tex]: is the initial speed of the truck = 17.9 m/s
[tex] a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2} [/tex]
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
[tex] \mu = \frac{a}{g} [/tex]
[tex] \mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}} [/tex]
[tex] \mu = 0.35 [/tex]
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting point, a radar gun measures its speed as 50 mph. Assuming the car had a constant rate of acceleration, (a) calculate the time elapsed between when the car started at 10 mph to when its speed was measured and (b) what will the speed of the car be another 500 ft downstream of this point
Answers
Answer:
a) t = 11.2 s
b) v = 70.5 mph
Explanation:
a)
Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:[tex]t = \frac{v_{f} - v_{o}}{a} (1)[/tex]
where vf = 50 mph, and v₀ = 10 mph.However, we still lack the value of a.Assuming that the acceleration is constant, we can use the following kinematic equation:[tex]v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x (2)[/tex]
Since we know that Δx = 500 ft, we could solve (2) for a.In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:[tex]v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s (3)[/tex]
[tex]v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s (4)[/tex]
We can do the same process with Δx, from ft to m, as follows:[tex]\Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m (5)[/tex]
Replacing (3), (4), and (5) in (2) and solving for a, we get:[tex]a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} = \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m} = 1.6 m/s2 (6)[/tex]
Replacing (6) in (1) we finally get the value of the time t:[tex]t = \frac{v_{f} - v_{o}}{a} = \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2} = 11.2 s (7)[/tex]
b)
Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:[tex]v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)[/tex]
If we convert vf again to mph, we have:[tex]v_{f} = 31.5m/s*\frac{1mi}{1609m} *\frac{3600s}{1h} = 70.5 mph (9)[/tex]
A 0.15 kg baseball collides with a 1.0 kg bat. The ball has a velocity of 40 m/s immediately before the collision. The center of mass of the bat also has a velocity of 40 m/s, but in the opposite direction, just before the collision. The coefficient of restitution between the bat and the ball is 0.50. Estimate how fast the baseball is moving as it leaves the bat following the collision.
Answers
Answer:
The final velocity of the baseball as it leaves the bat is 40 m/s
Explanation:
The given parameters of the baseball and bat are;
The mass of the baseball = 0.15 kg
The mass of the bat = 1.0 kg
The velocity of the ball before collision, v₁ = 40 m/s
The velocity of the bat before collision, v₂ = -40 m/s
The coefficient of restitution, e = 0.50
Let, 'v₃', and 'v₄' represent the final velocity of the ball and the bat respectively after collision, we have;
Taking the final velocity of the bat, v₄ = 0 m/s
According to Newton's Law of restitution
e = (v₃ - v₄)/(v₁ - v₂)
∴ 0.5 = (v₃ - v₄)/(40 - (-40))
80 × 0.5 = 40 = (v₃ - v₄)
v₃ - v₄ = 40
v₃ = 40 + v₄ = 40 + 0 = 40
The final velocity of the baseball as it leaves the bat, v₃ = 40 m/s.
If a shopping cart is pushed by a person exerting 50 J of work on it, what is the energy transfer to the shopping cart if it has a mass of 2 kg?
A. 52 J
B. 50 J
C. 25 J
Answers
Answer:
B
Explanation:
m
A 3.0 kg model train going right at 2.8 bumps into another 2.0 kg model train car moving in the same
S
m
direction at 1.6 . The heavier train car has a final speed of 2.2 to the right.
S
S
What is the final speed of the lighter 2.0 kg train car?
Answers
Answer:
it’s 2.5 m/s
Explanation:
i’m too lazy but trust
This question can be solved by using the law of conservation of momentum.
The final speed of the lighter 2 kg train is " 2.5 ".
When two moving objects collide with each other, the law of conservation of momentum can be applied to them as follows:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of heavier train = 3 kg
m₂ = mass of lighter train = 2 kg
u₁ = initial speed of heavier train = 2.8
u₂ = initial speed of lighter train = 1.6
v₁ = final speed of heavier train = 2.2
v₂ = final speed of lighter train = ?
Therefore,
(3 kg)(2.8) + (2 kg)(1.6) = (3 kg)(2.2) + (2 kg)(v₂)
[tex]v_2 = \frac{5 kg}{2 kg}[/tex]
v₂ = 2.5
Learn more about the law of conservation of momentum here:
https://brainly.com/question/1113396?referrer=searchResults
The attached picture illustrates the law of conservation of momentum.
Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir level dropped by 0.75 m in spite of an average inflow into the reservoir of 0.5 Mm3 /day. During the month, the average seepage loss from the reservoir was 2.5 cm, total precipitation on the reservoir was 18.5 cm and the total evaporation was 9.5 cm
Answers
Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s
The rate of withdrawal is 10.11 cubic meter per second
Given-
Total reservoir is 1375 hectare. which is equal to [tex]1375\times 10^4[/tex] meter square.
The average seepage loss from the reservoir is 2.85 cm or 0.0285 m.
Total precipitation on the reservoir is 18.5 cm or 0.185 m.
Total evaporation is 9.5 cm or 0.085 m.
The average inflow into the reservoir is [tex]0.5\times10^6[/tex] cubic meter per day.
The total inflow in a month can be calculate is
[tex]=0.5\times 30=1500\times10^4[/tex]
Net inflow is equal to the total precipitation on the reservoir subtract by all the losses.It can be represent as,
[tex]Q_{net}=0.185 - 0.095 - 0.025[/tex]
[tex]Q_{net}=0.065[/tex]
Total volume inflow is equal to the product of net inflow and total reservoir,
[tex]V_{net} =1375\times 10^4\times Q_{net}[/tex]
[tex]V_{net} =1375\times 10^4\times 0.065[/tex]
[tex]V_{net} =89.375[/tex]
The constant rate of withdrawal in cubic meter can be calculated by adding the net inflow in a month, total volume inflow and the reservoir.
[tex]Q=1375\times 10^4\times 0.75+89.375\times 10^4+1500\times10^4[/tex]
[tex]Q=2620\times10^4[/tex]
For per second withdrawal,
[tex]Q=\dfrac{2620\times10^4}{30\times24\times60\times60}[/tex]
[tex]Q=10.11[/tex]
Hence, the rate of withdrawal is 10.11 cubic meter per second.
For more about the flow, follow the link below,
https://brainly.com/question/1410288
Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 6.17-kg crate to the bottom of a steep ravine of height 23.8 meters. The 55.6-kg crewman is walking along holding the rope, being careful to lower the crate at a constant speed of 1.50 m/s. Unfortunately, when the crate reaches a point 13.2 meters above the ground, the crewman steps on a slick patch of ice and slips. The crate immediately accelerates toward the ground, dragging the hapless crewman across the ice and toward the edge of the cliff.
If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), at what speed will the crate hit the ground? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff. At what speed will the crewman hit the bottom of the ravine?
Answers
Answer:
a. Vc = 5.06 m/s
b. Vp = 22.18 m/s
Explanation:
The acceleration of the pulley-mass system is as follows:
a = [tex]\frac{mg}{m + M}[/tex]
Solving for acceleration, we get:
a = [tex]\frac{6.17 *9.8}{6.17 + 55.6}[/tex]
a = 0.97
So, for the part a:
Calculate the velocity of the crewman by using the following equation:
Vc = [tex]\sqrt{Vi^{2} + 2ay}[/tex]
Substituting the values into the equation, we get:
Vc = [tex]\sqrt{1.50^{2} + 2*0.97*13.2}[/tex]
Vc = 5.06 m/s
Now, for part b:
Calculate the final velocity of the pulley by using the following expression:
Vp = [tex]\sqrt{Vi^{2} +2gy }[/tex]
Just plugging in the values.
Vp = [tex]\sqrt{5.06^{2} +2*9.8*23.8 }[/tex]
Vp = 22.18 m/s
A hedgehog lives in a backyard in England. Every night, the house owner puts out a bowl of canned cat food and hard-boiled egg. The hungry hedgehog eats some of the food, then stops when it is no longer hungry. This pattern helps the hedgehog to maintain a steady energy level and weight.
What is the name for keeping a stable internal environment despite changes in the outside environment?
A. hormonal control
B. stimulus and response
C. balance
D. homeostasis
Answers
Answer: I think it C and B but I am really confident in C
Explanation:
YALL PLEASE HELP I BARELY HAVE TIME
Which of the following is not a property of light?
Light travels in a straight line.
Light travels through empty space.
Light moves in a compressional wave.
All options are true
Answers
Answer:
All of then are true
I need brainliest so I can rank up
Explanation:
Answer:
I think all options are true is the right answer
Explanation:
Mark me the brainliest plzzz
Plzz help me with this
I’ll give brainliest
Answers
Answer:
B. Objects with more mass have more gravitational force acting upon them.
Answer:
Should be A but it can be B as well.
A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.2 seconds, coasts for 2.5 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. How far apart are the stop signs
Answers
Answer:
the distance between the stop signs is 120.7 m.
Explanation:
The car moved in three stages;
(1) It accelerates from rest at 2.0 m/s² for 6.2 seconds
(2) it moved at a constant speed for 2.5 s
(3) it finally decelerate at the rate of 1.5m/s²
(1) The distance moved by the car during the first stage;
s₁ = ut + ¹/₂at²
s₁ = 0 + ¹/₂ (2)(6.2)²
s₁ = 38.44 m
(2) The distance moved by the car during the second stage;
calculate the constant speed of the car,
v = u + at
v = 0 + 2 x 6.2
v = 12.4 m/s
The distance moved by the car as it coasts for 2.5s: s₂ = vt
s₂ = 12.4 x 2.5
s₂ = 31 m
(3) The distance moved by the car during the third stage;
When the car stops, the final velocity is zero.
v² = u² + 2as₃
a = -1.5 m/s², since the car slowed down or decelerated.
0 = 12.4² + (2 x - 1.5)s₃
0 = 153.76 - 3s₃
3s₃ = 153.76
s₃ = 153.76 / 3
s₃ = 51.253 m
The total distance moved by the car from the start to stop = s₁ + s₂ + s₃
d = 38.44 m + 31 m + 51.253 m
d = 120.7 m
Therefore, the distance between the stop signs is 120.7 m.
The steepness of a line on a graph is called the
O A. rise
OB. slope
C.
run
D. verticle axis
Answers
Answer:
slope
Explanation:
The slope is how how steep the line is.
