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3 years ago

6

Suppose a distant world with surface gravity of 5.20 m/s2 has an atmospheric pressure of 7.16 ✕ 104 Pa at the surface. (a) What

force is exerted by the atmosphere on a disk-shaped region 2.00 m in radius at the surface of a methane ocean

1 answer:

Nesterboy [21]3 years ago

6 0

Answer:

899752.13598 N

Explanation:

p = Pressure on the plate = $7.16\times 10^4\ Pa$

A = Area = $\pi r^2=\pi 2^2=\pi 4$

F = Force on the region

When force is divided by area we get pressure

$p=\frac{F}{A}\\\Rightarrow F=p\times A\\\Rightarrow F=7.16\times 10^4\times \pi 4\\\Rightarrow F=899752.13598\ N$

Force is exerted by the atmosphere on a disk-shaped region is 899752.13598 N

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kherson [118]

Answer:

The largest longitude separation is 3.6 cm.

Explanation:

Given that,

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Wide = 10 cm

Vertical shear = 1000 N

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Using formula of shear flow

$q=\dfrac{VQ}{I}$

Where, q = shear flow

V = shear force

$Q = \bar{y}\times A$

Where, A = area of cross section

$\bar{y}$ =distance from natural axis to centroid of A

I = moment of inertia

We need to calculate the Area

Using formula of area

$A=4\times10=40 cm^2$

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$I=\dfrac{bd^3}{12}$

$I=\dfrac{10\times12^3}{12}$

$I=1440\ cm^4$

Put the value into the formula of shear flow

$q=\dfrac{1000\times40\times4}{1440}$

$q=111.11\ N/cm$

We need to calculate the largest longitude separation

Using formula of separation

$q=\dfrac{F}{\delta}$

Put the value into the formula

$111.11=\dfrac{2\times200}{\delta}$

$\delta=\dfrac{2\times200}{111.11}$

$\delta=3.6\ cm$

Hence, The largest longitude separation is 3.6 cm.

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4 years ago

A parallel-plate capacitor with circular plates of radius 0.19 m is being discharged. A circular loop of radius 0.28 m is concen

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Answer:

$\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s$

Explanation:

Given that

R= 0.19 m

r= 0.28 m

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We know that

$I_D=\epsilon _oA\dfrac{dE}{dt}$

A= Area of loop

dE/dt= rate of change of electric filed

I=Displacement current

Here r>R

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Now by putting the values

$I=\epsilon _oA\dfrac{dE}{dt}$

$2.6=8.85\times 10^{-12}\times \pi\times 0.19^2\dfrac{dE}{dt}$

$\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s$

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4 years ago

1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s ,

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Answer:

https://youtu.be/ymHHdoCGJOU

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3 years ago

a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the we

andreev551 [17]

Explanation:

The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.

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∑τ = Iα

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W = 200 N

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4 years ago

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Answer:

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