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2 years ago

The electric field inside a cell membrane is 8.0 MN/C. Part A What's the force on a singly charged ion in this field?

Physics

1 answer:

dlinn [17]2 years ago

7 0

Answer:

Force on the singly charged ion will be $12.6\times 10^{-13}N$

Explanation:

We have given electric field E = 8 MN/C

So electric field in N/C will be $E=8MN/C=8\times 10^6N/C$

It is given that ion so charge on ion will be equal to $e=1.6\times 10^{-19}C$

We have to find the electric force on the ion

Electric force is equal to $F=qE$ , here q is charge and E is electric field

So force on the charge will be equal to $F=8\times 10^6\times 1.6\times 10^{-19}=12.6\times 10^{-13}N$

So force on the singly charged ion will be $12.6\times 10^{-13}N$

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The speed v of a sound wave traveling in a medium that has bulk modulus b and mass density ρ (mass divided by the volume) is v=b

PilotLPTM [1.2K]

As it is given that Bulk modulus and density related to velocity of sound

$v = \sqrt{\frac{B}{\rho}}$

by rearranging the equation we can say

$B = \rho * v^2$

now we need to find the SI unit of Bulk modulus here

we can find it by plug in the units of density and speed here

$B = \frac{kg}{m^3} * (\frac{m}{s})^2$

so SI unit will be

$B = \frac{kg}{m* s^2}$

SO above is the SI unit of bulk Modulus

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2 years ago

Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti

san4es73 [151]

A) $1.36\cdot 10^{-4}T$

The magnetic field at the center of a coil of N turns is given by

$B=\frac{\mu_0 N I}{2R}$

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

$R=\frac{6.0 m}{2}=3.0 m$ is the radius of the coil

Substituting,

$B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T$

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) $3.26\cdot 10^{-23}N$

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

$F=qvB$

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

$q=+e=1.6\cdot 10^{-19} C$ is the charge

$v=1.5 m/s$ is the velocity

$B=1.36\cdot 10^{-4}T$ is the magnetic field strength

Substituting,

$F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N$

8 0

2 years ago

Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb

melomori [17]

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

$F=G\frac{Mm}{r^2}$

$F=G\frac{M6m}{r_b^2}$

$G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r$

$r_b=2.44948r$

8 0

2 years ago

A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d

arsen [322]

Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7 , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional magnetic flux is induced out of the page.

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2 years ago

A radioactive source has a half life of 80s.

Burka [1]

Lets make the original number of nuclides at the start is 100.

If 7/8 of 100 is decayed, that means 87.5 decayed.

$\frac{7}{8} \times 100 = 87.5$

And there is 1/8 left of the number of nuclide 100. Which is 12.5

$100 - 87.5 =12.5$

$\frac{1}{8} \times 100 = 12.5$

How many Half lifes passed for 100 to become 12.5 is 3 Half-Lives.

$100 \div 2 \div 2 \div 2 = 12.5$

Each Half-Life is 80 seconds so there is 240 seconds

$3 \times 80 = 240$ The answer is that it takes 240 seconds.

6 0

2 years ago