Michelle melted a compound in the lab and performed the same procedure five times. The chart below shows the time it took for the compound to melt during the five trials.Trial Time (min:sec)
1 3:14
2 3:10
3 4:05
4 3:12
5 3:09

After reviewing her data, she realized that trial 3 was vastly different from the other four trial times. Which of the following reasons could explain the radical difference in melting time for trial 3?

A larger quantity of compound was used in trial 3.
The compound started at a lower temperature in trial 3.
The burner was not directly under the compound during trial 3.
all of these

Answers

Answer 1
Answer:

Answer:

It is actually all of these

Explanation:

Answer 2
Answer:

Final answer:

Several factors could account for the variance seen in the third trial's melting time. This could be due to using a larger quantity of compound, starting at a lower temperature, or the burner not being directly under the compound.

Explanation:

The radical difference in melting time for trial 3 in Michelle's experiment could be explained by several variables. Given the information present, Michelle could have used a larger quantity of compound in trial 3, thus taking longer to melt. Secondly, the compound may have started at a lower temperature, requiring additional time to reach its melting point. Finally, if the burner was not directly under the compound, this could also explain the longer time as the heat wouldn't be as concentrated on the compound. In other words, the answer could be 'all of these' as they are all potential reasons as to why the third trial had a heavier melting time.

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Direct current is the flow of an electric charge _____.A. that regularly reverses direction B. that regularly stops and starts again C. that moves in only one direction D. that stops at a specific point
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Some vehicles, like _____, have different rules for registration and renewal. A. trailers B. coupes C. sedans D. pickup trucks

Answers

Some vehicles, like trailers, have different rules for registration and renewal

What is the registration process for trailers?

Before registering a trailer, you will generally need to submit a signed title to verify your ownership. Alternately, you might be able to use a manufacturer’s statement of origin.

You will also need to sign and submit an application for the license and title of the trailer, along with receipts of applicable property taxes. Finally, you will need to pay the registration fees.

Before you make a trip to the DMV, make sure to have two forms of photo ID on hand to verify your identity when you speak with an agent. For example, your driver’s license might serve as the primary form of ID, and a passport could serve as the second.

If you do not have a passport, you could use a military identification card instead. If your only form of photo ID is your driver’s license, be sure to have other forms of identification on hand, such as your birth certificate and the card with your Social Security number.

Hence Some vehicles, like trailers, have different rules for registration and renewal.

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Answer:

trailers

Explanation:

Within an internal combustion engine, the can-shaped component that moves up and down the cylinder and is the main moving part in the engine is the A. connecting rod.
B. crankshaft.
C. piston.
D. spark plug.

Answers

Answer;

C. piston.

Explanation;

  • A piston is a cylindrical component of the engine that slides back and forth in the cylinder bore by forces produced during the combustion process. It is the can shaped engine component that moves up and down the cylinder and  is the main moving part of an engine.
  • It acts as a movable end of the combustion chamber. Pistons are made of cast aluminium alloy to allow excellent and light weight thermal conductivity. They consists of a piston head, piston pin bore, piston pin, skirt piston rings, ring grooves and ring lands.
The crankshaft turns that moves the connecting rod up and down and the piston is inside the cylinder up and down movement.Answer piston

What affect does doubling the net force have on the acceleration of the object (whenthe mass of the object stays the same)? Identify a set of two trials that support your answer to question 1

Answers

Answer: The acceleration doubles

===========================================================

Explanation:

Consider a mass of 10 kg, so m = 10

Let's say we apply a net force of 20 newtons, so F = 20

The acceleration 'a' is...

F = ma

20 = 10a

20/10 = a

2 = a

a = 2

The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.

---------------

Now let's double the net force on the object

F = 20 goes to F = 40

m = 10 stays the same

F = ma

40 = 10a

10a = 40

a = 40/10

a = 4

The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.

---------------

In summary, if you double the net force applied to the object, then the acceleration doubles as well.

Acceleration is directly proportional to the net force on an object, and inversely proportional to its mass.

So if an object's mass stays the same while the net force on it doubles, then its acceleration will also double.

We don't know anything about the "trials".  This sounds like it might be a follow-up to a lab experiment that was performed when we weren't there.

We also don't know anything about "question 1".

Sound waves in air are a series ofa. periodic disturbances.
b. periodic condensations and rarefactions.
c. high- and low-pressure regions.
d. all of the above E. none of the above

Answers

Sound waves in air are a series of periodic disturbances, periodic condensations and rarefactions, and high- and low-pressure regions. It is all of the above. The answer is letter D.

Final answer:

Sound waves in air are a series of periodic condensations and rarefactions.

Explanation:

Sound waves in air are a series of periodic condensations and rarefactions. When a sound is produced, it creates compression regions (areas of high pressure) and rarefaction regions (areas of low pressure) that propagate through the air. These alternating regions of high and low pressure create the characteristic waveform of the sound wave.

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Two cars collide at an intersection. Car A, with a mass of 1900 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 17.0 m\s. As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 60.0degrees south of east from the point of impact.Part A WAS: How fast were the enmeshed cars moving just after the collision? I got 8.66 for velocity in part a which was CORRECT but i can't figure out PART B??...Part B:How fast was car A going just before the collision

Answers

Part A: The enmeshed cars were moving at a velocity of approximately 8.66 m/s just after the collision.

Part B: Car A was traveling at a velocity of approximately 8.55 m/s just before the collision.

How to compute the above velocities

To find the speed of car A just before the collision in Part B, you can use the principle of conservation of momentum.

The total momentum of the system before the collision should equal the total momentum after the collision. You already know the total momentum after the collision from Part A, and now you want to find the velocity of car A just before the collision.

Let's denote:

- v_A as the initial velocity of car A before the collision.

- v_B as the initial velocity of car B before the collision.

In Part A, you found that the enmeshed cars were moving at a velocity of 8.66 m/s at an angle of 60 degrees south of east. You can split this velocity into its eastward and southward components. The eastward component of this velocity is:

v_east = 8.66 m/s * cos(60 degrees)

Now, you can use the conservation of momentum to set up an equation:

Total initial momentum = Total final momentum

(mass_A * v_A) + (mass_B * v_B) = (mass_A + mass_B) * 8.66 m/s (the final velocity you found in Part A)

Plug in the known values:

(1900 kg * v_A) + (1500 kg * v_B) = (1900 kg + 1500 kg) * 8.66 m/s

Now, you can solve for v_A:

(1900 kg * v_A) + (1500 kg * v_B) = 3400 kg * 8.66 m/s

1900 kg * v_A = 3400 kg * 8.66 m/s - 1500 kg * v_B

v_A = (3400 kg * 8.66 m/s - 1500 kg * v_B) / 1900 kg

Now, plug in the values from Part A to find v_A:

v_A = (3400 kg * 8.66 m/s - 1500 kg * 8.66 m/s) / 1900 kg

v_A = (29244 kg*m/s - 12990 kg*m/s) / 1900 kg

v_A = 16254 kg*m/s / 1900 kg

v_A ≈ 8.55 m/s

So, car A was going at approximately 8.55 m/s just before the collision in Part B.

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The car salesman tells you that the car can go from a stopped position to 60 miles per hour in 6 seconds. He is giving you the car's rate of

Answers

he is giving you the car rate speed