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3 years ago

Why does lithostatic pressure increase more rapidly with depth than hydrostatic pressure?

1 answer:

Rudiy273 years ago

3 0

<span>Answer: Because rock is more dense than water

Lithostatic pressure is pressure caused by soil or rock. It similar to hydrostatic pressure, the pressure caused by water. Pressure is force per area and force is dependent on the mass. Rock is denser than water, thus having more mass and more force.</span>

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A balance uses small metal weights of known mass to determine the mass of a substance. Where would a balance NOT function correc

tatuchka [14]

Answer:

D is the answer.

Explanation:

Just do it.

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3 years ago

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The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of

PolarNik [594]

The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions

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3 years ago

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A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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3 years ago

Electric power companies measure energy consumption in kilowatt-hours, denoted kWh. One kilowatt-hour is the amount of energy tr

Natali [406]

The amount of energy used in the billing period is 5,400,000,000 joules.

One kWH is the amount of energy transferred in one hour, there 1 kWh is equal to

1 kWh=1*1000*(J/s)*3600 s

=3600000 J

Thus the amount of energy in joules consumed by the user for the billing period is =1500*3600000=5,400,000,000 J.

The amount of energy used in the billing period is 5,400,000,000 joules.

7 0

3 years ago