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3 years ago

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The diagram below shows a portion of the rock cycle.

1 answer:

SCORPION-xisa [38]3 years ago

8 0

Referring to the attached image below which represents the complete rock cycle, we will find that a marble can be formed when hot magma is cooled and crystallized.

Applying this to the given diagram, we will find that marble is most likely formed in position C

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Convert 3.45inches into km

Katyanochek1 [597]

Answer:

can someone please answer this i need this for a mastery test aswell

Explanation:

it would be very appreciated

8 0

3 years ago

Read 2 more answers

A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Pig. P7.41). (a) How fast

Leona [35]

Answer:

a)Velocity of car =v=16 m/s

b)Force against the track at point B=1.15* $10^{4}$ N

Explanation:

Given mass of roller coaster=m=350 kg

Position of A=Ha=25 m

Position of B=Hb=12 m

Net potential energy=mg(ha-hb)

Net potential energy=(350)(9.80)(25-12)

Net potential energy=44590 J

Using energy conservation

net kinetic energy=net potential energy

(1/2)mv^2=mg(ha-hb)

m=350

velocity=v=16 m/s

b)There two force acting,centripetal force upward and gravity downward.

Thus net force acting will be

Net force=(mv^2/r)-mg

Net force=14933.33-3430

Net force=1.15* $10^{4}$ N

3 0

3 years ago

Read 2 more answers

A sprinter starts from rest and runs with constant acceleration to a top speed of 12m/s in 4.0 seconds.

bagirrra123 [75]

Answer:

a = 3,0 m/s²

Explanation:

En este ejercicio se pide calcular la aceleracion del cuerpo, usemos las ecuaciones de cinematica en una dimensión.

v= v₀ + a t

como el corredor parte del reposo si velocidad inicial es cero

v = at

a = v/t

calculemos

a = 12 /4,0

a = 3,0 m/s²

4 0

3 years ago

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

natka813 [3]

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

8 0

3 years ago

A 18-g paper clip is attached to the rim of a phonograph record with a diameter of 48 cm, spinning at 3.2 rad/s. What is the mag

Artist 52 [7]

Answer:

Magnitude of its angular momentum = 0.0017 kgm²/s

Explanation:

Angular momentum, L = Iω

I is mass moment of inertia and ω is angular velocity.

Phonograph is in disc shape,

$\texttt{I for disc =}\frac{MR^2}{2}$

Radius = 0.5 x 48 = 24 cm = 0.24 m

Angular velocity, ω = 3.2 rad/s

Mass, M = 18 g = 0.018 kg

Substituting

$L=\frac{0.018\times 0.24^2}{2}\times 3.2 =0.0017kgm^2/s$

Magnitude of its angular momentum = 0.0017 kgm²/s

5 0

3 years ago