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tamaranim1 [39]

2 years ago

15

Earlier we discussed the concept of isostasy, where lower density rocks rise higher than higher density rocks. How is the variat

ion of water depth at spreading centers (ridges) controlled by isostasy

1 answer:

Elena L [17]2 years ago

4 0

The variation of water depth at spreading centers (ridges) controlled by isostasy as convective cooling cools the rocks much more effectively the than heat conduction.

<h3>What is convective heat transfer?</h3>

When heat transfer takes place between the two fluids in direct or indirect contact.

The lithosphere cools when it moves away from the ridge axis by sea floor spreading. The cooler rocks have low density, so the sea floor gets deeper as the lithosphere gets more dense.

Thus, the convective cooling cools the rocks much more effectively the than heat conduction.

Learn more about convective heat transfer

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billiardballsnew2 A white billiard ball with mass mw = 1.49 kg is moving directly to the right with a speed of v = 3.09 m/s and

Kobotan [32]

Answer:

Final velocity of white ball is 0m/s

Final velocity of black ball is 3.09m/s

Explanation:

An elastic collision is one that conserves internal kinetic energy

An internal kinetic energy is the sum of kinetic energies of objects in the system

Initial kinetic energy of white ball is Vi1 = 3.09m/s

Final kinetic energy of white ball is Vf1 = ?

Initial kinetic energy of black ball is Vi2 = 0m/s

Final kinetic energy of black ball is Vf2 = ?

m1 = 1.49kg mass of white ball

m2 = 1.49kg mass of black ball

The formula to calculate internal kinetic energy is

1/2m1Vf1^2 + 1/2m2Vf2^2 = 1/2m1Vi1^2

Solving the equation

1.Vf1 = (m1 - m2)Vi1/m1+m2

Vf1 = (1.49-1.49)*3.09/1.49+1/49

Vf1 = 0m/s

2. Vf2 = 2m1Vi1/m1+m2

Vf2 = 2*1.49*3.09/1.49+1.49

Vf2 = 3.09m/s

N:B following the general principle of collision when 2 bodies of same masses collide in elastic collision they exchange velocities.

7 0

4 years ago

The spring is compressed a total of 3.0 cm, and used to set a 500 gram cart into motion. Find the speed of the cart at the insta

BartSMP [9]

Answer:

1.15 m/s

Explanation:

Part of the question is missing. Found the missing part on google:

"1. A hanging mass of 1500 grams compresses a spring 2.0 cm. Find the spring constant in N/m."

Solution:

First of all, we need to find the spring constant. We can use Hooke's law:

$F=kx$

where

$F=mg=(1.5 kg)(9.8 m/s^2)=14.7 N$ is the force applied to the spring (the weight of the hanging mass)

x = 2.0 cm = 0.02 m is the compression of the spring

Solving for k, we find the spring constant:

$k=\frac{F}{x}=\frac{14.7}{0.02}=735 N/m$

In the second part of the problem, the spring is compressed by

x = 3.0 cm = 0.03 m

So the elastic potential energy of the spring is

$U=\frac{1}{2}kx^2=\frac{1}{2}(735)(0.03)^2=0.33 J$

This energy is entirely converted into kinetic energy of the cart, which is:

$U=K=\frac{1}{2}mv^2$

where

m = 500 g = 0.5 kg is the mass of the cart

v is its speed

Solving for v,

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.33)}{0.5}}=1.15 m/s$

4 0

3 years ago

Convert 3.8 kg to Lbs

JulijaS [17]

I believe it’s 8.3 to be precise

8 0

2 years ago

Read 2 more answers

For the strongest electromagnet, what wire gauge should be used?

alexandr1967 [171]

Answer:

Thick

Explanation:

Thicker wure enables higher AMP's, assuming your supply is capable, so indirectly thicker wire may help. Thicker wire takes up more space. So does more turns.

Hope this helped. Have a blessed day !

7 0

2 years ago

A 1500-kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200-kg SUV traveling from ea

astra-53 [7]

Answer:

Velocity of Sedan = 21 m/s

Velocity of SUV = 12 m/s

Explanation:

As we know that deceleration due to friction force is given as

$a = - \mu g$

so we have

$a = -(0.75)(9.81)$

$a = -7.36 m/s^2$

now the two cars comes to rest at a point which is at position of 5.39 m West and 6.43 m South

so net displacement of the car is given as

$d = \sqrt{5.39^2 + 6.43^2}$

$d = 8.39 m$

now the velocity of the two cars just after the impact is given as

$v^2 - v_i^2 = 2a d$

$0 - v_i^2 = 2(-7.36)(8.39)$

$v_i = 11.11 m/s$

direction of the motion is given as

$tan\theta = \frac{6.43}{5.39}$

$\theta = 50 degree$ South of West

now we can use momentum conservation as there is no external force on it

Momentum conservation in North to south direction

$m_1 v_1 = (m_1 + m_2) vsin\theta$

$1500 v_1 = (1500 + 2200) (11.11) sin50$

$v_1 = 21 m/s$

Similarly momentum conservation towards West direction

$m_2 v_2 = (m_1 + m_2) vsin\theta$

$2200 v_2 = (1500 + 2200) (11.11) cos50$

$v_2 = 12 m/s$

5 0

3 years ago