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3 years ago

A train travels 81 kilometers in 2 hours and then 90 kilometers in two hours. wat is the average speed

1 answer:

5 0

I'm assuming your question refers to the train travels 81 km in 2 hours, as in that's the total distance covered versus the speed for those 2 hours, whereas the 90 km in 2 hours the second time around is the total distance not speed I would assume.

Now... if it was going 81 km for the first 2 hours and 90 km for the second 2 hours then the average speed would be the mean of these numbers, with that being 85.5 km. Though, I doubt that's your question.

With that said, 81 km covered by 2 hours and 90 km covered by 2 more hours. To acquire the km an hour average, we'll have to divide the distance by how many hours it traveled:
81 / 2 = 40.5
90 / 2 = 45

Meaning the train was going 40.5 km an hour for the first 2 hours and 45 km an hour for the second 2 hours. Now, to find the mean:
40.5 + 45 = 85.5 / 2 = 42.75

In-case you were wondering, the mean is the sum of all the numbers in a set, divided by the total amount of numbers in that set, take for instance:
3, 5, 9, 2, 1, 5. -> There are 6 numbers in this set.
3 + 5 + 9 + 2 + 1 + 5 = 25 -> 25 is the sum of these numbers.
25 / 6 = 4.2 (estimated) -> 4.2 is roughly the mean or average between the original set.

Anyhow, with that aside the average speed between this I would believe would be 42.75 km an hour. I hope that helps, have a great rest of your day! ^ ^
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A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.

Luba_88 [7]

Answer: Option (b) is the correct answer.

Explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

F = $\frac{kqQ}{r^{2}} - kq\frac{(2Q)}{(2r)^{2}}$

= $\frac{4KqQ - 2KqQ}{4r^{2}}$

= $\frac{KqQ}{2r^{2}}$

F = $\frac{KqQ}{2r^{2}}$ > 0

Thus, we can conclude that the test charge move to the right immediately after being released.

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3 years ago

A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc

OLga [1]

Answer:

5 N

Explanation:

The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N

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3 years ago

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Goals can be big or small? Question 1 options: True False

pav-90 [236]

Answer:

Probably big

Explanation:

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2 years ago

Why is this event important to include in the biography?

nadya68 [22]

Answer:

D: It shows that Frida Kahlo used art to cope with her pain.

Explanation:

Within the text given it shows her emotions being lonely, immobile and in pain. But it all shows her asking her father for art which states that art is her sort of relief and happy place.

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2 years ago

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I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

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