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3 years ago

Which feature forms at a divergent boundary

Physics

2 answers:

Romashka-Z-Leto [24]3 years ago

4 0

They normally form new land or ocean trenches.

Ber [7]3 years ago

4 0

Answer:

Rift valleys

Explanation:

Divergent boundaries are formed when the tectonic plates move away from each other forming rift valleys. Often volcanic islands are also formed. These divergent boundaries are found at the base of oceans.

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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

Irina18 [472]

Answer:

a. $q_C$ experiences the greatest net force and $q_B$ experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

<u>Electrostatic Forces </u>

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

$\displaystyle F=K\frac{q_1q_2}{d^2}$

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

$F_{TA}=-F_B-F_C+F_D$

Computing the individual forces we have

$\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}$

$\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}$

$\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}$

The total force on qA is:

$\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}$

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

$\displaystyle F_{TB}=F_A-F_C+F_D$

$\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}$

$\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

$\displaystyle F_{TC}=F_A+F_B+F_D$

$\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}$

$\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

$\displaystyle F_{TD}=-F_A-F_B-F_C$

$\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}$

$\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}$

Comparing the relative values of all the forces

$\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|$

This means that qc experiences the greatest net force and qB experiences the smallest net force

The ratio of the greatest to the smallest forces is

$\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9$

5 0

3 years ago

Find the frequency, if the amplitude of a 3000g object in simple harmonic motion is 1000cm and the maximum speed of the object i

Oksana_A [137]

Answer:

A = 10 m amplitude

m = 3 kg mass of object

Vm = 5 m/s

w A = Vm where w = omega

w = 2 * pi * f

2 * pi * f 10 = 5

f = 5 / (20 * pi) = .0796 / sec

7 0

2 years ago

What change of state occurs during vaporization? liquid to gas gas to liquid solid to liquid liquid to solid

nignag [31]

Answer:

A is correct!

Explanation:

e2020

8 0

3 years ago

Read 2 more answers

Help me please

agasfer [191]

Answer:

For a velocity versus time graph how do you know what the velocity is at a certain time?

Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.

_____________________________________________________

How do you know the acceleration at a certain time?

$Ans: We\ know\ that\ acceleration = \frac{Final\ velocity-Initial\ Velocity}{Time\ taken}$

Hence,

By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.

_________________________________________________________

How do you know the Displacement at a certain time?

Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.

_________________________________________________________

4 0

3 years ago

Mr. Beall paddles his canoe at 8.0 m/s North in a river that flows at 6.0 m/s to the South. What is the magnitude and direction

kifflom [539]

Answer:

2 /s north

Explanation:

Given that,

Velocity due North is 8 m/s and due south is 6 m/s

We need to find the magnitude and the direction of the resulting velocity.

Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

$v=8+(-6)\\\\v=2\ m/s$

It is positive. So, it is in North direction.

3 0

3 years ago