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3 years ago

What is the strength of the electric field inside the membrane just before the action potential?

Physics

1 answer:

gtnhenbr [62]3 years ago

5 0

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

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How long will it take a person walking at 2.1 m/s to travel 13 m?

MrRissso [65]

Answer:

I gonna give you the number so but you need to round 6.19047619048

Explanation:

This is a speed formula so you would use the formula speed=distance/time
You need to rearrange it to time=distance/speed
So you need to divide 13m by 2.1 m/s

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3 years ago

Which law states that that the direction of the induced current is such that the magnetic field resulting from the induced curre

Vikentia [17]

Lenz's law or option B for plato users

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3 years ago

Read 2 more answers

A 13,000 kg helicopter accelerates upward a 0.5 m/s^2 while lifting a 2000 pound car. to the nearest newton, what is the lift fo

Vinil7 [7]

Lift force exerted by the air on the rotors=143244 N

Explanation:

we use Newtons second law

F- (M+m)g=(M+m)a

F= lift force

m= mass of helicopter= 13000 Kg

M= mass of car= 2000 lb=907.2 kg

a= acceleration= 0.5 m/s²

g= acceleration due to gravity

F- (M+m)g=(M+m)a

F=(M+m)(a+g)

F=(13000+907.2)(0.5+9.8)

F=143244 N

8 0

3 years ago

A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta

serg [7]

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

$mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s$

So, the final velocity of the bullet is 9 m/s.

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3 years ago

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

igomit [66]

Answer:

Angular acceleration of the disk will be $\alpha =10.714rad/sec^2$

Explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius $r=\frac{d}{2}=\frac{0.3}{2}=0.15m$

Moment of inertia of disk is given by $I=\frac{1}{2}mr^2=\frac{1}{2}\times 5\times 0.15^2=0.056kgm^2$

Force is given by F=4 N

Torque is given as $\tau =Fr=4\times 0.15=0.6N-m$

We also know that torque is given by $\tau =I\alpha$

$0.6=0.056\times \alpha$

$\alpha =10.714rad/sec^2$

5 0

3 years ago