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adelina 88 [10]

3 years ago

7

Which has the least gravitational potential energy?

1 answer:

kogti [31]3 years ago

8 0

Answer: it’s a 10kg barbell resting on the floor

Explanation:

This has the least gravitational potential energy

You might be interested in

Methane gas reacts with oxygen in a gas stove.

nekit [7.7K]

Hey there!

CH₄ + 2O₂ → CO₂ + 2H₂O

carbon dioxide: product

methane: reactant

oxygen: reactant

water: product

Methane and oxygen are reactants because they are the substances we start with. They are on the side of the equation that the arrow is pointing away from.

Carbon dioxide and water are products because they are the new substances that are the yields of the equation. They are on the side of the equation that the arrow is pointing towards.

Hope this helps!

7 0

3 years ago

Costco installs automobile tires on a first-come first-serve basis. the total time a customer needs to wait for the installation

den301095 [7]

There you go that’s the answer

3 0

3 years ago

Write a balanced chemical equation for the formation of magnesium oxide, mg0, from magnesium and oxygen

swat32

This picture explains:

7 0

3 years ago

Where are people mostly exposed to chemicals?

Vlad [161]

Answer:

people are mostly exposed to chemicals through their nose, mouth, eyes, and ears

Explanation:

they are the easiest way for anything, bad or good, to enter the body because people touch their face alot.

This might not answer what you were looking for but the other answer covered it pretty well so I thought I'd give you another angle to the question!

Hope this helps!!

4 0

3 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

4 years ago