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Aleks [24]
3 years ago
13

Describe the role of the plasma membrane

Physics
1 answer:
lara31 [8.8K]3 years ago
5 0
The plasma membrane of a cell is a group of lipids and proteins that forms the boundary between a cell's contents and the outside of the cell.
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What do we measure sound intensity in?
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we measure sound intensity in <em><u>D</u></em><em><u>ecibels</u></em>.

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The light reactions of photosynthesis supply the calvin cycle with __________.
Margaret [11]

Answer:

ATP and NADPH

Explanation:

4 0
3 years ago
Which of your groups should you NOT change anything for? (in other
ella [17]

Answer:

B

Explanation:

The control is something that is meant to not be changed, the control is a comparison of the experimental.

7 0
3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
4 years ago
A car is traveling at 21m/s. It accelerates at 1.4m/s2 for 11s. How fast is the car moving after the acceleration?
nata0808 [166]

Answer:

v=u+at

v=21+1.4(11)

v=21+15.4

v=36.4

3 0
3 years ago
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