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3 years ago

8

What causes the pressure that allows diamonds to form in the mantle

2 answers:

LuckyWell [14K]3 years ago

5 0

It’s C I just took it

icang [17]3 years ago

3 0

Answer:

C. Downward force from rocks in the crust

Explanation:

Diamonds are different forms of element carbon. Diamonds are formed in Mantle and they require large amount of temperature and pressure for their formation. These come up in higher layers due to volcanic eruptions. The pressure is caused by the downward force from the rocks in the crust. The pressure is not constant everywhere because the thickness of the layer varies. Thus, diamonds are not that common.

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A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0

3 years ago

The weight of the atmosphere above 1 m- of

mars1129 [50]

Answer:

1.09 kg.m

Explanation:

no need

4 0

2 years ago

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$

$\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)$

$\Delta E=170418.5\ J$

$\Delta E=1.7\times 10^5\ J$

(b) Change in momentum of the truck, $\Delta p=m(v-u)$

$\Delta p=2000\ kg\times (16.11-9.44)$

$\Delta p=13340\ kg-m/s$

Hence, this is the required solution.

6 0

3 years ago

Problem set 3 » (8) river swimming a swimmer heads directly across a river, swimming at her maximum speed of 1.30 m/s relative t

jasenka [17]

Velocity of swimmer across river = 1.30 m/s

Distance arrived downstream = 48 m

Width of river = 64 m

Time taken to cross river = $\frac{Width of river}{Velocity across river}$

= $\frac{64}{1.30} =49.23 s$

Speed of river current = $\frac{Distance arrived downstream}{Time taken to cross river}$

= $\frac{48}{49.23} = 0.975 m/s$

So, the river is flowing at a speed 0.975 m/s.

5 0

3 years ago

The number of wave cycles that pass in one second is called the _______ of the wave.

postnew [5]

That is the frequency.

6 0

3 years ago

Read 2 more answers