Ask question

Ask question

All categories

3 years ago

14

17. In which layer does mantle convection occur?

1 answer:

Anarel [89]3 years ago

8 0

Answer:

D. Asthenosphere

Explanation:

The asthenosphere is relatively plastic part of the mantle which underlies the brittle lithosphere. In the asthenosphere, it is generally believed that the rocks are in ductile state and easily moves. It is the site of convection within the earth. In mantle convection, hot and light materials rises and keeps moving into upper crustal levels till they solidify. Here also, cold and denser materials sinks deeper till they turn to melt. This differences in temperature and density sets up a convective cell within the mantle. Several convective cells are in the mantle.

You might be interested in

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

2 years ago

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago

How much physical activity should an adult have each week?

LuckyWell [14K]

Answer:d

Explanation:

7 0

2 years ago

Read 2 more answers

40 Points!!!!!!!!!!!!!

Firlakuza [10]

The answer is A for sure

3 0

2 years ago

The length and mass of the arm are Larm = X1 = 50 cm and Marm = 0.3 kg, X2 = 15 cm, and the mass of the object is MObject = 0.25

max2010maxim [7]

Answer: 0.5N

Explanation: if the system is at equilibrium, sum of the torque will be equal to zero.

But if they are not in equilibrium.

U will find the difference in the two torque

find the attached file for solution

3 0

2 years ago