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RoseWind [281]
3 years ago
14

17. In which layer does mantle convection occur?

Physics
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.

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Find the value of F1 + F2 + F3.<br>​
Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u />F_{y}=2[N]<u />

<u>For F₂</u>

F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]

<u>For F₃</u>

<u />F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]<u />

Now we can sum each one of the forces in the given axes:

F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]

Now using the Pythagorean theorem we can find the total force.

F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]

8 0
3 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

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We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

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3 years ago
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A bicyclist is travelling at 25 m/s when he begins to decelerate at -4m/s2 . How fast is travelling after 5 seconds
Umnica [9.8K]

Initial velocity (Vi) = 25 m/s

acceleration (a) = -4 m/s^{2}

time interval (t) = 5 sec

let us assume that final velocity after 5 sec be Vf

As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e. V_{f} = V_{i} + at

Hence, V_{f} = 25 +(-4)(5) = 25 -20 = 5 m/s

so, the velocity of bicyclist will be 5 m/s after 5 sec

7 0
3 years ago
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