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snow_lady [41]
3 years ago
9

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

e a parallel plate capacitor, each with a charge density of 10?5C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.
PART A. What is the magnitude of the electric field between the membranes?

1×106N/C

1×10?15N/C

5×10?5N/C

9×10?2N/C
Physics
1 answer:
Nina [5.8K]3 years ago
4 0

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

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Which form of energy does the motor convert most of its electric energy into?
aev [14]

Answer:

Kinetic energy

Explanation:

6 0
2 years ago
Read 2 more answers
A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio
Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

  • A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.
  • Treat them separately.
  • At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.
  • Horizontal movement is not influenced by gravity.
  • acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

d_x=10.0m\\d_y=3.0m

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ?     Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

d_y=V_i_yt+\dfrac{1}{2}at^{2}

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

d_y=0+\dfrac{1}{2}at^{2}

d_y=\dfrac{1}{2}at^{2}

From here we can derive the formula of time:

t=\sqrt{\dfrac{2d_y}{a}}

Now we just plug in what we know:

t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

vf_y=vi_y+at

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em>d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x</em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17

The lion leapt at an angle of 50.16°.

6 0
2 years ago
A pendulum is transported from sea-level, where the acceleration due to gravity g is 9.80 m/s2, to the bottom of Death Valley. A
tatyana61 [14]

Answer: the value of g in Death Valley is 10.417 m/s²

Explanation:

Given that;

acceleration due to gravity at the point is g = 9.8 m/s²

Lets say the acceleration due to gravity at the bottom of Death valley is g'

as the period of the pendulum is decreased by 3.00%

T' = 0.97 T  

T is the period of the pendulum at sea level  and T' is the period of the pendulum at bottom of Death valley

therefore from the relation

T = 2π√(l/g)

g'/g = T²/T'²

g' = (T²/ (0.97T)²)g

g' = 1.063g

g' = 10.417 m/s²

therefore the value of g in Death Valley is 10.417 m/s²

7 0
3 years ago
________ can occasionally cause the North Atlantic to appear green and murky.
castortr0y [4]

Answer:

algae

Explanation:

The Atlantic off the East Coast of the United States usually appears green. This is due to the presence of algae and plant life.

4 0
2 years ago
Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher
siniylev [52]

Answer:

v = 4.44 \times 10^5 m/s

Explanation:

By Einstein's Equation of photoelectric effect we know that

h\nu = W + \frac{1}{2}mv^2

here we know that

h\nu = energy of the photons incident on the metal

W = minimum energy required to remove photons from metal

\frac{1}{2}mv^2 = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

W = \frac{hc}{351 nm}

here we know that

hc = 1242 eV-nm

now we have

W = \frac{1242}{351} = 3.54 eV

now by energy equation above when photon of 303 nm incident on the surface

\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2

4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2

(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2

8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2

v = 4.44 \times 10^5 m/s

6 0
3 years ago
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