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ch4aika [34]
3 years ago
5

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

igher potential than the outer surface. A potassium ion (K+) is initially just outside the cell membrane (initially at rest). How much work (in J) is required for a cell to absorb the ion, so that it moves from the exterior of the cell to the interior?
Physics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer: W = 1.04.10^{-20} J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

W=q.\Delta V

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = 1.6.10^{-19} C

To determine work in joules, potential has to be in Volts, so:

\Delta V=65.10^{-3}V

Then, work is

W=1.6.10^{-19}.65.10^{-3}

W=1.04.10^{-20}

To move a potassium ion from the exterior to the interior of the cell, it is required W=1.04.10^{-20}J of energy.

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(b) The particle displacement y of air molecules due to a sound wave is given by y 4m and w = = 0.008 cos wt sin kz. Where k - m
serious [3.7K]

The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.

<h3>What's the distance between consecutive nodes of the displacement of air molecules?</h3>
  • Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.
  • So, distance between consecutive nodes = wavelength = 2π÷k

= 2π/(4π÷m)

= m/2

<h3>What's the amplitude after 0.56s of the displacement of air molecules?</h3>

Displacement after 0.56 s = 0.008×cos(50π×0.56s)

=1.75×10^(-4) m

Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.

Calculate:

I) the distance between 2 consecutive nodes

ii) the amplitude after 0.565s

Learn more about the wavelength here:

brainly.com/question/10750459

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8 0
1 year ago
2. Use physics terms to explain the benefits of crumple zones in modern cars.
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Answer:

They decrease trauma by allowing for a more gradual change in velocity

Explanation:

6 0
3 years ago
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

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3 years ago
Two forces act on an object. The first force has a magnitude of 15.0 N and is oriented 78.0 ° counterclockwise from the + x ‑axi
bezimeni [28]

Answer:

c

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The average specific heat of the human body is 3.6 kJ/kg·°C. If the body temperature of a(n) 96-kg man rises from 37°C to 39°C d
d1i1m1o1n [39]

Answer:

691200 J

Explanation:

From specific heat capacity,

ΔQ = cmΔt.................. Equation 1

Where ΔQ = increase in thermal energy, c = specific heat capacity of the body, m = mass of the man, Δt = rise in temperature.

Given: c = 3.6 kJ/kg.°C = 3600 J/kg.°C, m = 96 kg, Δt = 39-37 = 2 °C.

Substitute into equation 1

ΔQ = 3600×96×2

ΔQ = 691200 J.

Hence the change in the thermal energy of the body = 691200 J

5 0
3 years ago
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