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3 years ago

What causes the pressure to increase as you go deeper below the crust?

Physics

2 answers:

Masteriza [31]3 years ago

8 0

The deeper you go, the more rock must be supported so the more force is required and the pressure goes up.

enyata [817]3 years ago

7 0

Answer:

Pressure is force per unit area. In this case the force is primarily the weight of the overlying rock. So , in very simple terms, the deeper you go the more rock must be supported so the more force is required and the pressure goes up. It is basically the same thing that happens in the ocean and in the atmosphere.

Explanation:

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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

An airplane flies 120 km at a constant altitude in a direction 30.0° north of east. a wind is blowing that results in a net hori

astraxan [27]

Best Answer:<span> </span><span>The net angle between the direction of flight of the aircraft and the wind in the opposing direction is 20. THe component opposing the aircraft is 2.4*cos 20 KN
= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km
The work done by the aircraft overcoming the wind is
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N
As the question is trickily worded as : the work done on the plane by the air (wind) the answer is -2.71 x 10^8 J . (fourth option)</span>

6 0

3 years ago

Read 2 more answers

Martha must carry a 45 N package up three flights of stairs. Each flight of stairs has a height of 2.3m, and the actual distance

Darina [25.2K]

Answer:

310.5 J

Explanation:

The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to

$\Delta U = mg\Delta h$

where

(mg) = 45 N is the weight of the package

$\Delta h$ is the increase in height of the package

The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is

$\Delta h = 3 \cdot 2.3 m=6.9 m$

And so, the work done by Martha is

$U=(45 N)(6.9 m)=310.5 J$

3 0

3 years ago

If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?

Tamiku [17]

The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

$R=A=B$

So, using the law of cosines, we have:

$R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°$

Hence, we have that the angle between the vectors is 120°. The correct option is C) The angle between the vectors is 120°

Have a nice day!

4 0

3 years ago

31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r

posledela

Answer:

See answers below

Explanation:

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

7 0

2 years ago