Why does a spherometer have three legs

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

Nina [5.8K]

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

4 0

3 years ago

Which is true about the way air flows

JulsSmile [24]

Answer:

A High-to-Low

Explanation:

its like water running down a hill.

4 0

3 years ago

If you were going to describe the relationship between current, voltage, and power, you could say: (1 point) Group of answer cho

Ber [7]

Answer:

The correct option is "In order to gain more power you would need to increase either current or voltage."

Explanation:

To answer the question, we note that;

The formula for Electrical Power are as follows,

P = I²·R, or P = I·V,

Therefore, if we increase either the current, I with the voltage, V remaining constant or we increase the Voltage, V with the current, I remaining constant or we increase both the voltage, V an the current, I the Power, P will be increased.

Therefore, the correct option is "In order to gain more power you would need to increase either current or voltage."

3 0

3 years ago

A solid, horizontal cylinder of mass 10.6 kg and radius 1.00 m rotates with an angular speed of 8.00 rad/s about a fixed vertica

n200080 [17]

Answer:

The final angular speed is 7.71 rad/s

Explanation:

Given

Cylinder mass, M = 10.6 kg

Cylinder radius, R = 1.00 m

Angular speed, w = 8.00 rad/s.

Mass of putty, m = 0.250-kg

Radius, r = 0.900 m

First, we set up an expression for the initial and final angular momentum of the system.

The moment of inertia of the cylinder is given as I = ½MR²

While the moment of inertia if the putty is mr².

Initial Momentum of the system = Initial momentum of the cylinder =

Li = Iw --- Substitute ½MR² for I

Li = ½MR²w

By

Substituton

Li = ½ * 10.6 * 1² * 8

Li = 42.4kgm²/s

Calculating the final momentum of the system.

First we calculate the final momentum of the cylinder

Li = Iw --- Substitute ½MR² for I

Li = ½MR²wf where wf = final angular speed

By

Substituton

Li = ½ * 10.6 * 1² wf

Li = 5.3w kgm²/s

Then we calculate the final momentum of the putty

Final Momentum of the putty =

L2 = Iwf --- Substitute mr² for I;

L2 = mr²wf --- By Substituton

L2 = 0.25 * 0.9² * wf

L2 = 0.2025wf kgm²/s

Final momentum = Li + L2

Lf = (5.3wf + 0.2025wf) kgm²/s

Lf = 5.5025wf kgm²/s

By conservation of momentum

Li = Lf

Where Li = 42.4kgm²/s and Lf = 5.5025wf kgm²/s

So, we have

5.5025wf kgm²/s = 42.4kgm²/s --- make wf the subject of formula

wf = 42.4/5.5025

wf = 7.71 rad/s

Hence, the final angular speed is 7.71 rad/s

5 0

3 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

Now the distance between the charge at at origin and the second charge:

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

Now the distance between the charge at at origin and the third charge:

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

Now the force due to second charge:

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

Now the force due to third charge:

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

Now the its horizontal component:

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

Now the its vertical component:

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

3 0

3 years ago

Why does a spherometer have three legs​

Why does a spherometer have three legs