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rjkz [21]
3 years ago
15

Individual M has longer limbs and muscles than Individual N. Both individuals lift the same amount of weight from the ground to

their shoulders an equal number of times. If all repetitions were completed in synchrony, which statement about the amount of work completed by the individuals is true? At the end of the workout:
A. Individual M will have done less work because the object was moved a greater distance
B. both individuals will have done the same amount of work because force was applied for the same duration of time.
C. Individual M will have done more work because the muscle contracted faster to achieve synchrony with Individual N
D. Individual N would have done less work because the object was moved a shorter distance
Chemistry
1 answer:
Elena L [17]3 years ago
5 0

Answer:

D. Individual N would have done less work because the object was moved a shorter distance

Explanation:

When a body is moved a given distance (x) by a specific force (F), the work done on the body is equivalent to F*x. This shows that work is proportional to the applied force and the distance traveled by the body. Therefore, the higher the distance the higher the work done and the lower the distance the lower the work done. The work done by Individual N is lower due to the shorter limbs that correspond to shorter distance.

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represent rutheniums first 36 electrons using the chemical symbol for krypton written inside the brackets
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[kr] hd^75s^1

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Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(N
Rina8888 [55]

Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

7 0
3 years ago
Se construye una pila galvánica con una barra de cobre sumergida en una disolución 1M de cationes Fe+2 y una barra de plata sume
pashok25 [27]

Answer:

El potencial celular estándar, E_{cell} is +0.46 V

Explanation:

Las reacciones de media célula son;

Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V

Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V

Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V

y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V

 que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo

Por lo tanto, en el ánodo, tendremos

Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)

En el cátodo

2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)

E_{cell} = E_c + E_a = -0.34 + 0.8 = +0.46 \, V

El potencial celular estándar, E_{cell} = +0.46 V

4 0
3 years ago
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