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mina [271]
3 years ago
14

The voltage across a membrane forming a cell wall is 84.0 mV and the membrane is 9.40 nm thick. What is the electric field stren

gth (in V/m)
Physics
1 answer:
Katarina [22]3 years ago
7 0

Answer:

8.9*10^6 V/m

Explanation:

The expression for electric field strength E is given as

E=\frac{V}{d}

where V= voltage

           d= distance of separation

Given data

voltage= 84 mV= 84*10^-^3\\distance= 9.4*10^-^9

substituting our given data into the electric field strength formula we have

E= \frac{84*10^-^3}{ 9.4*10^-^9} \\\\E= \frac{84}{9.4} *10^-^3^-^(^-^9^)\\\\E= \frac{84}{9.4} *10^-^3^+^9\\\\E= \frac{84}{9.4} *10^6\\\\E=8.9*10^6  V/m

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The electric field strength is 20,000 N/C inside a parallel-platecapacitor with a 1.0 mm spacing. An electron is released fromre
creativ13 [48]

Explanation:

It is given that,

Electric field strength, E = 20,000 N/C

Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m

Initial velocity of electron, u = 0

Let v is the electron’s speed when it reaches the positive plate. The force acting on the electron is :

F=qE

Also, ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\times 20,000}{9.1\times 10^{-31}}

a=3.51\times 10^{15}\ m/s^2

Using third equation of motion as :

v^2-u^2=2ad

v=\sqrt{2ad}

v=\sqrt{2\times 3.51\times 10^{15}\times 0.001}

v = 2649528.2599 m/s

or

v=2.64\times 10^6\ m/s

So, the velocity of the electron when it reaches the positive plate is 2.64\times 10^6\ m/s. Hence, this is the required solution.

4 0
3 years ago
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A girl is riding her bicycle. She rides 4 miles in 2.5 hours. What was her average speed?
xz_007 [3.2K]

Explanation:

4/2.5 = 1.6 miles/hour

4 0
3 years ago
A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(
maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

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3 years ago
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Stolb23 [73]

Answer:

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3 years ago
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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

Magnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9  N

4 0
4 years ago
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