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2 years ago

11

What is the rate at which work is done?

1 answer:

LiRa [457]2 years ago

5 0

Answer:

The rate at which work is done would be power. Bonus: The SI unit of work is Joules.

Explanation:

Hope this helped, nya~ :3

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When removing the objects from the oven, you accidentally touch each one with your hand. Rank these objects on the basis of how

kykrilka [37]

Answer:

So, we can assemble the options based or evaluation of there properties regarding getting an equilibrium or balanced state in a given time.We have the following rankings attributed to the elements:

Silveringpot≥ aluminiumpot ≥ironskillet ≥glasscasseroldish ≥welldone steak ≥woodencuttingboard.

Explanation:

<u>Attaining equilibrium matters:</u>

When the materials are placed inside the oven, they attain a high temperature value causing it to be non touchable but some of the items has low value to attain the equilibrium state when comes in contact with other mediums. As these materials are also arranged based on that analyses.

8 0

3 years ago

The law of conversation of energy states that energy cannot be created or

irinina [24]

Answer:

It says energy can't be created or destroyed

Explanation:

5 0

3 years ago

If a rock is dropped from the top of a tower at the front of it and takes 3.6 seconds to hit the ground. Calculate the final vel

expeople1 [14]

Answer:

35.28m/s; 63.50m

Explanation:

<u>Given the following data;</u>

Time, t = 3.6 secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the final velocity, we would use the first equation of motion;

$V = u + at$

Substituting into the equation, we have;

$V = 0 + 9.8 * 3.6$

V = 35.28m/s

Therefore, the final velocity of the penny is 35.28m/s.

To find the height, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting the values into the equation;

$S = 0(3.6) + \frac {1}{2}*9.8*(3.6)^{2}$

$S = 0 + 4.9*12.86$

$S = 0.5 *36$

S = 63.50m

Therefore, the height of the tower is 63.50m.

6 0

3 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0

3 years ago

If we have less power, we most likely have

boyakko [2]

The same amount of work being done over a long period of time!

6 0

3 years ago