Ask question

Ask question

All categories

4 years ago

9

A cell membrane is 0,0000005 m thick and has an electric potential difference between its surfaces of 0,10 V. What is the electr

ic field within the membrane?

1 answer:

Irina-Kira [14]4 years ago

5 0

Answer:

Electric field in the membrane is given as

$E = 2 \times 10^5 V/m$

Explanation:

As we know that electric field in a region is related to its potential difference given by the formula

$E = \frac{\Delta V}{\Delta x}$

here we know that

$\Delta V = 0.10 V$

$\Delta x = 0.0000005 m$

so we have

$E = \frac{0.10}{0.0000005}$

$E = 2 \times 10^5 V/m$

You might be interested in

A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if

tigry1 [53]

Answer:

4.16m/s²

Explanation:

According to Newtons second law;

$\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\$

Fm is the moving force

$\mu$ is the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is 4.16m/s²

4 0

3 years ago

A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here <u>friction force on the given block will be same as its component on weight which is 218.55 N</u>

5 0

3 years ago

A 2200 kg car starts from rest and speeds up to 12 m/s in 5.2 s. The net force acting on the car is?

zmey [24]

The acceleration is 12(m/s)/5.2s = 2.308m/s^2
F=ma, so
F=2200*2.308 = 5076.9N

6 0

3 years ago

A cylinder of radius r=10.0 cm and height h=20.0 cm

Olenka [21]

D I think is the correct answer
If the cylinder is slightly

7 0

2 years ago

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

4 years ago