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3 years ago

Explain the purpose of a geologic time scale

2 answers:

7 0

Geologic time scales are essential tools for geologists and biologists alike because they help show the history of the entire world in terms of relationships between different plate movements.

sladkih [1.3K]3 years ago

6 0

Geologic time scales are essential tools for geologists and biologists alike as a result of they assist show the history of the whole world in terms of relationships between totally different plate movements.

Explanation:

Time could be an important variable in earth science, as a result of the exact timing of spatially separated events permits the United States of America to reconstruct the surface and surface conditions of the ancient earth. ... geological time spans area unit significantly more difficult to grasp than historical time spans as a result of they're therefore incredibly long.

In the geological timescale, time is usually divided on the premise of the earth's biotic composition, with the Phanerozoic aeon(i.e. the era, Mesozoic era and cenozoic Eras) representing the amount of Earth's history with advanced life forms, and therefore the Pre Cambrian (or Proterozoic and Hadean Eras.

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A given wave has a wavelength of 5.0 m and a frequency of 2.0 Hz. How fast

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B is the answer 3.0 m/a

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3 years ago

Which leader maintained control of Cuba from 1934 to 1959?

Alexeev081 [22]

The leader Fulgencio Batista maintained control of Cuba from 1934 to 1959.

Answer: B. Fulgencio Batista

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4 years ago

Why is the work output of a machine never equal to the work input? A. Some work input is used to overcome friction. B. Input dis

SVETLANKA909090 [29]

The answer is A. Some work input is used to overcome friction.

6 0

3 years ago

Read 2 more answers

A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig

inna [77]

Answer:

$\mu_s=\frac{1}{3}\tan \theta$

Explanation:

Let the minimum coefficient of static friction be $\mu_s$ .

Given:

Mass of the cylinder = $M$

Radius of the cylinder = $R$

Length of the cylinder = $L$

Angle of inclination = $\theta$

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

$\tau =I\alpha$

Now, angular acceleration is given as:

$\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}$

Also, moment of inertia for a cylinder is given as:

$I=\frac{MR^2}{2}$

Therefore, the torque acting on the cylinder can be rewritten as:

$\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1$

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are $mg\sin \theta\ and\ f$ . The net force acting along the incline is given as:

$F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2$

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, $N=Mg\cos \theta$

Plugging in $N=Mg\cos \theta$ in equation (2), we get

$F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3$

Now, as per Newton's second law,

$F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4$

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

$\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)$

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

$\mu_s=\frac{1}{3}\tan \theta$

3 0

3 years ago

Read 2 more answers

Can you make the work output of a machine greater than the work input?

ExtremeBDS [4]

Answer:

It is impossible to construct a machine which produces the work output greater than the work input.

Let us consider the II law of thermodynamics.

According to Kelvin Plank's statement any engine/machine does not give hundred percent efficiency. And violating the PMM-II(Perpetual motion of machine II kind), Always some amount of energy transferred to the sink or surroundings.

Therefore

W(ouput) = Q₁-Q₂

There are many reasons to lower the work output, just for an example friction between the mating parts reduces the work output.

5 0

3 years ago

Read 2 more answers