All categories

3 years ago

Help me!!!

Physics

1 answer:

Elena-2011 [213]3 years ago

8 0

Answer:

n a static situation the membrane has a charge distribution of −2.5 × 10 −6 C/m 2 on its inner surface and +2.5 × 10 −6 C/m 2 on its outer surface. Draw a diagram of the cell and the surrounding cell membrane. Include on this diagram the charge distribution and the corresponding electric field.

You might be interested in

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

magnetic field intensity, $B=0.07\ T$

kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

Now, form the mathematical expression of Kinetic Energy:

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

from the relation of magnetic and centripetal forces we have the radius as:

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

6 0

3 years ago

Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?

Mademuasel [1]

The first opiton is the answer A)Rahul’s weight

8 0

3 years ago

Read 2 more answers

What role does team collaboration play in successfully planning a mission to Mars?

mihalych1998 [28]

Answer:it helps get everyone in the same page

Explanation:

This is important so everyone knows what they should be doing

4 0

3 years ago

Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the g

Vladimir79 [104]

Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

v₂ = 306.12 m/s

5 0

3 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago