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3 years ago

A golf pro has 2100 J of kinetic energy and swings his driver which weighs .75 kg. What is the speed of his swing?

Chemistry

2 answers:

jeka57 [31]3 years ago

8 0

Answer : The correct answer is 74.83 m/s .

The kinetic energy is energy possessed by any mass which is moving or have some speed . It is product of mass and velocity . It is expressed as :

$KE = \frac{1}{2} * m* v^2$

Where KE = kinetic energy in J or $kg\frac{m^2}{s^2}$ m = mass in Kg v = speed in m/s²

Unit of KE is Joules (J) .

Givne : KE = 2100 J mass = 0.75 kg v = ?

Plugging value in KE formula =>

$2100 J = \frac{1}{2} * 0.75 Kg * v^2$

$2100 J = 0.375 Kg * v^2$

Dividing both side by 0.375 kg =>

$\frac{2100 J}{0.375 Kg} = \frac{0.375 Kg}{0.375 Kg } * v^2$

$v^2 = 5600 \frac{m^2}{s^2}$

$v= 74.83 \frac{m}{s}$

Kazeer [188]3 years ago

7 0

The speed of his swing is 56 m/s

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Katena32 [7]

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3 years ago

Which half-reaction correctly represents oxidation?

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Explanation:

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3 years ago

How many moles of CO gas are in 34.6 L?

Troyanec [42]

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<h3>Further explanation</h3>

Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions

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2 years ago

In the gold foil experiment, what was observed

natima [27]

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3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$