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3 years ago

The first step in turning a rock into a sediment is ________. Select one:

Physics

1 answer:

Debora [2.8K]3 years ago

6 0

Answer:

The first step in turning a rock into a sediment is Compaction .

Explanation:

Lithification is a process of changing rock into sediments. There are two steps for a rock to lithify. These steps are as follows

The first step of lithification is compaction where sediments are erosed together by weight of the it. Thus, the upper layers of sediments causes compaction of lower layers.
The next process of lithification is cementation. In this fluids fill the space between the loose particles.

Hence, the first step for turning rock into sediments is compaction.

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Sound that reaches the ears after bouncing off a wall or a floor is called _____.

andrew11 [14]

Answer:

Indirect sound

8 0

2 years ago

Read 2 more answers

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

3 0

3 years ago

The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor

mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

5 0

3 years ago

Read 2 more answers

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

DanielleElmas [232]

Answer:

$\alpha =-2.2669642\times^{-10}rad/s^2$

Explanation:

Angular acceleration is defined by $\alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}$

Angular velocity is related to the period by $\omega=\frac{2\pi}{T}$

Putting all together:

$\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})$

Taking our initial (i) point now and our final (f) point one year later, we would have:

$\Delta t=1\ year=(365)(24)(60)(60)s=31536000
s$

$T_i=0.0786s$

$T_f=0.0786s+7.03\times10^{-6}s$

So for our values we have:

$\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2$

Where the minus sign indicates it is decelerating.

8 0

3 years ago

What is the dependent variable in this

Whitepunk [10]

Answer:

"the plants had already started growing."

Explanation:

I think this is the answer because the definition of a dependent variable is the variable that is being affected by the change. Since the plants had already started growing BECAUSE of "They ran out of the farmer's compost, so

some of the plants received that compost

when the seeds were planted and other

plants got hardware store compost after

the plants had already started growing."

Sorry if I am wrong, I am just a 4th grader, pls don't hate on me, I am just trying to help :)

6 0

3 years ago