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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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Where should you place the values of the independent variable when constructing a data table?

kobusy [5.1K]

<span>On the y-axis (the bottom of the table) hope this helps</span>

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3 years ago

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If the mass of both weights is 150 gm, the first mass is located 25° north of east, the second mass is located 25° south of east

Pani-rosa [81]

Answer:

voltage is 1.38 V

Explanation:

given data

weights = 150 gm = 0.150 kg

sensitivity = 0.5 volts/Newton

to find out

large a voltage

solution

we know here in by symmetry so here in x axis

F3 will be

F3 = 2mgcos(20)

F3 = 2(0.150)(9.8)cos20

F3 = 2.76 N

so volatge is

voltage = 2.76×sensitivity

voltage = 2.76×0.5

voltage = 1.38

so voltage is 1.38 V

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4 years ago

An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the

Juliette [100K]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

$s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}$ (1)

Speeder

$s = s_{o} + v_{o,S}\cdot t$ (2)

Where:

$s_{o}$ - Initial position, measured in meters.

$s$ - Final position, measured in meters.

$v_{o,P}$ , $v_{o,S}$ - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

$a$ - Acceleration of the unmarked police car, measured in meters per square second.

$t$ - Time, measured in seconds.

$t'$ - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

$v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t$

$v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t$

$\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0$

If we know that $a = 1.6\,\frac{m}{s^{2}}$ , $v_{o,P} = 0\,\frac{m}{s}$ , $v_{o,S} = 53.4\,\frac{m}{s}$ and $t' = 2.2\,s$ , then we solve the resulting second order polynomial:

$0.8\cdot t^{2}-56.92\cdot t +3.872 = 0$ (3)

$t_{1} \approx 71.082\,s$ , $t_{2}\approx 0.068$

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

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3 years ago

1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of

nevsk [136]

Answer:

$4.8 \cdot 10^4 N$

Explanation:

The force of gravity acting on the satellite is given by:

$F=\frac{GMm}{r^2}$

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

$r=2.4\cdot 10^6 m$

Substituting into the equation, we find:

$F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N$

<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>

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3 years ago

Moving a spring toy up and down will create what kind of wave?

tia_tia [17]

I’m not a 100% sure but I believe the transverse wave

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3 years ago

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