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2 years ago

An electric field of 6.1x 10^5 N/C is directed along a charge 2.9 x 10^-19 C. What

Physics

1 answer:

Aloiza [94]2 years ago

4 0

Answer:

Electric field is a function 1/r^2

Explanation:

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The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ

Marizza181 [45]

The energy stored in the membrane is $6.44\cdot 10^{-14} J$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the material

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

$A=4.50\cdot 10^{-9} m^2$

$d=8.1\cdot 10^{-9} m$

Substituting, we find its capacitance:

$C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F$

Now we can find the energy stored: for a capacitor, it is given by

$U=\frac{1}{2}CV^2$

where

$C=2.26\cdot 10^{-11} F$ is the capacitance

$V=7.55\cdot 10^{-2} V$ is the potential difference

Substituting,

$U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J$

Learn more about capacitors:

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6 0

3 years ago

Sics and Chemistry A (2nd Six Weeks) - SC3200SFA/01

Andreyy89

Force = Mass * Acceleration therefore the red ball with the higher mass will have more force and greater acceleration

3 0

3 years ago

An arrow is moving at 35 m/s and travels for 5 seconds. how far did the arrow travel?

pashok25 [27]

The answer is 175 meters

8 0

3 years ago

1. Calculate the charge in a circuit which carries a current of 30 amperes for 5 seconds

sweet-ann [11.9K]

Answer:

1. 150C.

2. 50sec

3.1.5a

Explanation:

1. I = Q/T

Q= 30x5

=150c

2.applying the formulae, I = Q/T

T= Q/I

=500/10

=50sec.

3. using the formulae i=q/t

i= 120/80

=1.5a.

6 0

3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

2 years ago