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3 years ago

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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A planet moves in a circular orbit of radius 4.5x10^15 m with a period of 4

GaryK [48]

Answer:

2.6×10^{10} m/s

Explanation:

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3 years ago

The ratio of the inductive reactance to the

BARSIC [14]

Answer:

The non resonance frequency of the generator is = 1201.79 Hz

Explanation:

At resonance,

f₀ = 1/2π√LC..................... Equation 1

Where f₀ = resonance frequency, L = inductance, C = capacitance

making LC the subject of the equation

LC = 1/4πf₀²..................... Equation 2

Given: f₀ = 225 Hz, and π = 3.143

Substituting these values into equation 2,

LC = 1/(4×3.143²×225²)

LC = 1/2000385.9

LC = 5×10⁻⁷

If the ratio of capacitive reactance to inductive reactance = 5.36

1/2πfC/2πfL = 5.36

1/4π²f²LC = 5.36

Where f = frequency of the non resonant

making f the subject of the equation

f = 5.36/2π√LC ............. Equation 3

Substituting the value of LC = 5×10⁻⁷ into equation 3

f = 5.36/2×3.143√(5×10⁻⁷)

f = 5.36/(6.286×0.00071)

f = 5.36/0.00446

f = 1201.79 Hz

Thus the non resonant frequency of the generator is = 1201.79 Hz

6 0

2 years ago

In a stunt, three people jump off a platform and fall 8.5 m onto a large air bag. A fourth person at the other end of the air ba

docker41 [41]

Answer:

They Died Period NO MORE

8 0

3 years ago

a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio

Likurg_2 [28]

The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

The given parameters;

mass of the ball, m₁ = 0.8 kg
speed of the ball, u₁ = 2.5 m/s
mass of the object at rest, m₂ = 2.5 kg
final velocity of the object at rest, v₂ = 1 m/s

Let the final velocity of the 0.8 kg ball immediately after collision = v₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)

2 = 2.5 + (0.8)v₁

-0.5 = (0.8)v₁

$v_1 = \frac{-0.5}{0.8} \\\\v_1 = -0.625 \ m/s$

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

Learn more here: brainly.com/question/7694106

5 0

2 years ago

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

inysia [295]

Answer:

$v = 1,582 \ \frac{m}{s}$

Explanation:

We know that for circular motion the centripetal acceleration $a_c$ is:

$a_c = \frac{v^2}{r}$

where v is the speed and r is the radius.

The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

$a_c = 1.27 \frac{m}{s^2}$ .

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

$r = 1.7 * 10^6 \ m + 270 \ km$

$r = 1.7 * 10^6 \ m + 270 * 10^3 \ m$

$r = 1.7 * 10^6 \ m + 0.270 * 10^6 \ m$

$r = 1.97 * 10^6 \ m$

We can obtain the speed as:

$v^2 = a_c r$

$v = \sqrt{a_c r}$

$v = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}$

$v = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}$

$v = 1.582 \ 10^3 \ \frac{m}{s}$

$v = 1,582 \ \frac{m}{s}$

And this is the orbital speed.

7 0

3 years ago