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Morgarella [4.7K]
3 years ago
8

Which of the following best characterizes the free energy change ΔG for an endothermic reaction under physiological conditions?A

. The sign of ΔG is positive
B. The sign of ΔG is zero
C. The sign of ΔG is negative
D. The sign of ΔG cannot be determined without more information
Physics
1 answer:
mote1985 [20]3 years ago
6 0

Answer:

D

Explanation:

Whether endothermic or exothermic, for a reaction to be spontaneous, ∆G must be negative.

∆G=∆H-T∆S.

The values of temperature and entropy change can be determined when the values of enthalpy, entropy and temperature are known. When ∆G is zero, the reaction has attained equilibrium.

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What is the oscillation period of your eardrum when you are listening to the A4 note on a piano (frequency 440 Hz)?
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The period is the inverse of the frequence:

T=\dfrac{1}{f}=\dfrac{1}{440}=0.0022\overline{72}

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An airplane travels directly from Washington, D.C., to Atlanta, Georgia, a distance of 850 km at a velocity of 425 km/h southwes
Sloan [31]

Answer:

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Explanation:

4 0
3 years ago
A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is
Tpy6a [65]

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

  • Force applied = 100 N
  • Distance covered = 5 metres
  • Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

\bf \boxed{Work = Force \times displacement}

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

5 0
3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
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