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GREYUIT [131]
3 years ago
12

Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th

e potential inside the cell is − 85.5 mV and the potential outside the cell is zero. a. What is the change in the electrical potential energy of a single potassium ion as it moves across the membrane?b. Does the potential energy of the potassium ions increase, decrease, or remain the same as the ions move from the inside to the outside?
Physics
1 answer:
choli [55]3 years ago
3 0

Answer:

1.368\times 10^{-20}\ J

Explanation:

q = Charge in the potassium ion = 19e-18e

e = Charge of electron = 1.6\times 10^{-19}\ C

V_2-V_1 = Change in potential = 0-(-85.5\times 10^{-3})

Change in electric potential is given by

E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J

The energy is 1.368\times 10^{-20}\ J

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The manufacturer of a 6V dry-cell flashlight battery says that the battery will deliver 15 mA for 60 continuous hours. During th
Vaselesa [24]

Answer:

16200 J

Explanation:

t = Time the battery is on = 60 hours

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Average voltage

V=\dfrac{6+4}{2}=5\ V

Energy is given by

E=V\times I\times t

\\\Rightarrow E=5\times 15\times 10^{-3}\times 60\times 3600

\\\Rightarrow E=16200\ J

The energy delivered in the given time is 16200 J

6 0
3 years ago
Whose geocentric model was accepted for 1400 years
zysi [14]
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8 0
3 years ago
Riders in a carnival ride stand with their backs against the wall of a circular room of diameter
Veseljchak [2.6K]

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

                  = 45 \times \dfrac{2\pi}{60}

                  =4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

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now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

 μ = 0.11

hence, the correct answer  is option C

6 0
3 years ago
When an object is falling and reaches a constant velocity, the net force on the object is ____ and the weight of the object is e
Maslowich

When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>.  (choice-D)

5 0
2 years ago
Anyone know these questions?
salantis [7]
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Constant speed would mean that the two forces are equivalent
7 0
3 years ago
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