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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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tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

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3 years ago
The sun is the ___________ of all the energy that moves though food
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The sun is the <em><u>source</u></em> of all the energy that moves through food. It helps the plants to grow which in turn become food that we and other animals eat.

Hope it helps :) 

3 0
3 years ago
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The Hubble Space Telescope orbits 600 km above Earth's surface. Earth's radius is about 6370 km. Use the Pythagorean Theorem to
musickatia [10]

To solve this problem we will apply the concept given by Pythagoras in the description of the lengths of the legs of a rectangular triangle and if equality against the square of the hypotenuse, that is

a^2+b^2=c^2

Here,

a, b = Legs of a triangle

c = Hypotenuse

According to the attached chart then we would have to

a=x\\b = 6370km\\c = 600+6370km = 6970km

Substituting the given the lengths into the Pythagorean Theorem.

a^2+b^2 = c^2 \\x^2 +(6370)^2 = (6970)^2\\x = 2829.13km \approx 2830km

Therefore the distance x is 2830km.

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3 years ago
The opening to a cave is a tall, 30-cm-wide crack. A bat that is preparing to leave the cave emits a 30 kHz ultrasonic chirp. Ho
sweet-ann [11.9K]

Answer:

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Explanation:

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We considered the opening long and narrow, Using single slit diffraction formula:

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Considering m=1, The angle between first minimum from center of beam is:

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The width of beam is:

tanΘ=y/L

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Width=7.557 m

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GarryVolchara [31]

Answer:

A

Explanation:

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