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3 years ago

Which of the following energy changes occurs as a bow is drawn and released?

2 answers:

8 0

<span>3. elastic potential to kinetic
</span>Mechanical energy is the top suject while it has two types: Kinetic energy which is the energy in motion and potential energy which is the energy in reserve. The measure of both energy in motion and reserve is called Joules. Joules then is the International System of Measurement unit for energy, this is mainly used to account for scaling energy in all aspects.<span> </span>

Sonbull [250]3 years ago

6 0

Answer: The correct answer is 3.

Explanation: As the bow is drawn and released, it will experience a change in elastic potential energy to kinetic energy.

Elastic potential energy is defined as the amount of potential energy which is stored as a result of the deformation of the object which has some elasticity in it. For Example: Stretching of a spring, Drawing of a bow, etc..

Kinetic energy is the energy of the object which is possessed as a result of its motion.

When the bow is drawn, it first had some elastic potential energy and when it is released it had some kinetic energy in it.

Conclusion: The correct answer is 3.

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Radio station broadcast signals two different frequency bands. These are Called ?

rusak2 [61]

Answer:

I think it is AM and frequency

Explanation:

Sorry if i'm wrong ;)

6 0

3 years ago

Read 2 more answers

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

galina1969 [7]

Answer:

$\%\, yield\, \, SO_3=82.29$

Explanation:

First write the balance eqation of chemical reaction:

$2S(s) +3O_2(g) \rightarrow 2SO_3(g)$

Remember writing any chemical reaction from its elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.

mass of oxygen given=5gram

mole of oxygen $=5/32mol=0.16mol$

mass of sulpher given=6gram

mole of sulpher $=6/32mol=0.19mol$

from the above balanced equaion;

2 mole of sulphur reacts with 3 mole Oxygen completely

1 mole of sulphur reacts with 1.5 mole Oxygen completely

0.19 mole of sulphur reacts with $1.5\times 0.19$ i.e. 0.285 mole Oxygen completely.

but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent

so product will depend on the limiting reagent

from the balance equation

3 mole of sulpher gives 2 mole $SO_3$

1 mole of sulpher will give 2/3 mole $SO_3$

0.16 of sulpher will give 0.12 mole $SO_3$

mass of $SO_3$ =9.6gram this is theoreical production of $SO_3$

and

actual production of $SO_3$ =7.9gram

$\%\, yield \,\, SO_3=\frac{Actual \,yield}{theoretical\, yield}\times 100$

$\%\, yield\, \, SO_3=\frac{7.9}{9.6} \times 100=82.29$

$\%\, yield\, \, SO_3=82.29$

3 0

3 years ago

Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given

professor190 [17]

Answer: -227 kJ

Explanation:

The balanced chemical reaction is,

$C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]$

$-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]$

$\Delta H_{C_2H_2}=-227kJ$

Therefore, the enthalpy change for $C_2H_2$ is -227 kJ.

7 0

3 years ago

What is the PEN number for Oxygen?

nataly862011 [7]

Answer:

Explanation:

g o o g l e d it

7 0

3 years ago

When 125.0 g of ethylene (C2H4) burns in 60.0 grams of oxygen to give carbon dioxide and water, how many grams of CO2 are formed

gizmo_the_mogwai [7]

Answer:

66 g of CO₂

Solution:

The Balance Chemical Reaction is as follow,

C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O

Or,

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)

Step 1: Find out the limiting reagent as;

According to Equation 1,

56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂

So,

125 g of C₂H₂ will react with = X g of O₂

Solving for X,

X = (125 g × 160 g) ÷ 56.1 g

X = 356.5 g of O₂

It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.

Step 2: Calculate Amount of CO₂ produced as;

According to Equation 1,

160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂

So,

60.0 g of O₂ will produce = X g of CO₂

Solving for X,

X = (60.0 g × 176 g) ÷ 160 g

X = 66 g of CO₂

8 0

3 years ago