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makkiz [27]
3 years ago
10

Elias serves a volleyball at a velocity of 16 m/s. The mass of the volleyball is 0.27 kg. What is the height of the volleyball a

bove the gym floor when its total mechanical energy is 41.70 J? Round to the nearest tenth. m
Chemistry
2 answers:
frosja888 [35]3 years ago
6 0

Answer:

2,7 m

Explanation:

You can solve this doing an energy balance:

m*g*h-\frac{1}{2} *m*v^{2} =41,7[J]

Solving this equation to get h:

\frac{41,7- \frac{1}{2} *m*v^{2} }{m*g}=h

Replacing the values and solving brings to 2,7 m

inn [45]3 years ago
6 0

Answer:

Its 2.7m

Explanation:

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Looters break a statue into pieces. How do you expect the weathering of pieces of rock to change?
andreyandreev [35.5K]
The statue will weather faster because of more surface area.
5 0
3 years ago
What is the temperature of 0.5 moles of water vapor that occupies 120 dm3 and applies a pressure of 15,000 Pa to its container?
MArishka [77]
Use the ideal gas law:

PV = nRT

so, T = PV / nR

n=0.5
V= 120 dm^3 = 120 L  (1 dm^3 = 1 L)
R = 1/12
P = 15,000 Pa =  0.147 atm     (1 pa = 9.86 10^{-6} )

Put the values:

T = PV / nR
T = (0.147) (120) / (0.5) (1/12)
T= 426 K
6 0
3 years ago
Which element above has 8 protons in its nucleus?
Nataliya [291]
A. Oxygen. Oxygen has an atomic mass of 16. The atomic mass of an atom is the combined weight o the protons and neutrons. Since Oxygen's atomic mass is 16, it has 8 protons and 8 neutrons.
6 0
3 years ago
Read 2 more answers
An exhaled air bubble underwater at 290.
Vinil7 [7]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 290 kPa

Final pressure (P_{2}) = 104 kPa

Initial volume (V_{1}) = 18.9 ml

To find:

Final volume (V_{2})

We know;

From the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

   P × V = constant

   P ∝ \frac{1}{V}

From the above equation;

              \frac{P_{1} }{P_{2} }  = \frac{V_{2} }{V_{1} }

P_{1} represents the initial pressure of the gas

P_{2} represents the final pressure of the gas

V_{1} represents the initial volume of the gas

V_{2} represents the final volume of the gas

Substituting the values of the above equation;

                    \frac{290}{104} = \frac{V_{2} }{18.9}

             V_{2} = 52.7 ml

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

6 0
2 years ago
Calculate the DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. DH°f means delta or change
USPshnik [31]

Answer: +178.3 kJ

Explanation:

The chemical equation  follows:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]

We are given:

\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]

The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ

5 0
2 years ago
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