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Amanda [17]
3 years ago
8

True or false? the potential energy of a membrane potential comes solely from the difference in electrical charge across the mem

brane.
Physics
1 answer:
bazaltina [42]3 years ago
7 0
The statement would be False. T<span>he potential energy of a membrane potential comes solely from the difference in electrical charge across the membrane. In addition to that, membrane potential actually regulates the potential difference of nerve cells across the membrane estimated at 70 mV.</span>
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A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
rjkz [21]

Answer:

a) F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

b) \mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

\Sigma F_{y}=0

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

-F_{y}-W+N=0

When solving for the normal force we get:

N=F_{y}+W

we know that

W=mg

and

F_{y}=Fsin \theta

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

f_{k}=\mu_{k}N

so we can find the kinetic friction by substituting for N, so we get:

f_{k}=\mu_{k}(F sin \theta +mg)

Now we can find the sum of forces in x:

\Sigma F_{x}=0

so after analyzing the diagram we can build our sum of forces to be:

-f+F_{x}=0

we know that:

F_{x}=Fcos \theta

so we can substitute the equations we already have in the sum of forces on x so we get:

-\mu_{k}(F sin \theta +mg)+Fcos \theta=0

so now we can solve for the force, we start by distributing \mu_{k} so we get:

-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0

we add \mu_{k}mg to both sides so we get:

-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg

Nos we factor F so we get:

F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg

and now we divide both sides of the equation into (cos \theta-\mu_{k} sin \theta) so we get:

F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

f_{s}=\mu_{s}(F sin \theta +mg)

and

F cos θ = f

when substituting one into the other we get:

F cos \theta=\mu_{s}(F sin \theta +mg)

which can be solved for the static friction coefficient so we get:

\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

which is the answer to part b.

3 0
3 years ago
Read 2 more answers
The established value for the speed of light in a vacuum is 299,792,458 m/s. What is the order-of-magnitude of this number?
iogann1982 [59]
The order of magnitude is 10⁸ .
6 0
3 years ago
Which do you think will penetrate farther into a block of lead, x-rays, or gamma rays
disa [49]
Gamma rays because it has more penetrating power and frequency but shorter wavelength.
3 0
3 years ago
What does digital media allow you to do?
kifflom [539]
Digital media<span> are any </span>media<span> that are encoded in machine-readable formats. </span>
4 0
3 years ago
Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of
morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner t=\frac{295.6}{4.65}=63.56 s

time required by first runner to reach finish line=\frac{250}{3.1}=80.64 s

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0
3 years ago
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