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Zarrin [17]
3 years ago
15

A 0.080Kg robin, perched on a power line 6 m above the ground , swoops down to snatch a worm from the ground and then returns to

an 8 m high tree beach with his catch. By how much did the birds PE increase in his trip from the power line to the branch?
Physics
1 answer:
raketka [301]3 years ago
4 0
If the initial PE is mgh(initial) and the final PE is mgh(final) then mgh(initial)-mgh(final) should give you your change in PE.
So, without conversions your answer should be
[(.08)(9.8)(8)]-[(.08)(9.8)(6)]=change in PE which is 1.568Joules.
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A sample of nitrogen occupies 5.50 liters under a pressure of 900 torr at 25oC. At what temperature will it occupy 10.0 liters a
vitfil [10]

Answer:

<u>At 268.82°C</u> volume occupied by nitrogen is 10 liters at pressure of 900 torr.

Explanation:

Given:

Volume of a sample of nitrogen = 5.50 liters

Pressure = 900 torr

Temperature = 25°C

To find the temperature at which the nitrogen will occupy 10 liters volume at same pressure.

Solution:

Since the pressure is kept constant, so we can apply the temperature-volume law also called the Charles Law.

Charles Law states that the volume of a gas held at constant pressure is directly proportional to the temperature of the gas in Kelvin.

Thus, we have :

V ∝ T

\frac{V}{T}=k

where k is a constant.

For two samples of gases, the law can be given as:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

From the data given:

V_1=5.5\ l

T_1=25\ \°C =(273+25)K= 298 K

V_2=10\ l

We need to find T_2.

Plugging in values in the formula.

\frac{5.5}{298}=\frac{10}{T_2}

Multiplying both sides by T_2.

T_2\times\frac{5.5}{298}=\frac{10}{T_2}\times T_2

\frac{5.5}{298}T_2={10}

Multiplying both sides by \frac{298}{5.5}

\frac{298}{5.5}\times\frac{5.5}{298}T_2=\frac{10\times 298}{5.5}

T_2=541.82\ K

T_2=541.82\ K-273\ K = 268.82\°C

Thus, at 268.82°C volume occupied by nitrogen is 10 liters at pressure of 900 torr.

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The images below are models that represent the reactants and products of a chemical reaction.
Eva8 [605]

Explanation:

In a chemical reaction, when mass is conserved , the number of atoms or moles of the reactants must be equal to the number of moles or atoms in the products side.

From the diagram, we should carefully look to see if the number of atoms that makes up the reactants are equal to those on the product side.

 For example:

      A + B → AB

Here, mass is conserved because, on the reactant side, we have 1 atom of A and on the product side we have 1 atom of A

For B, on the reactant side, we have 1 atom of B and on the product side, we have 1 atom of B.

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D

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