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3 years ago

Please solve this question

Physics

1 answer:

lesantik [10]3 years ago

8 0

Answer:

88200 Pa

it is because

height =9m

density=1000kg/m(cube)

gravity = 9.8m/s(square)

now,

P=d×g×h

= 1000×9.8×9

=88200pa

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This is about the magnet fields. thanks in advance.

Wewaii [24]

The answer is A. The outer lines change as it moves

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2 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

7 0

3 years ago

A concert loudspeaker suspended high off the ground emits 31 W of sound power. A small microphone with a 1.0 cm^2 area is 42 m f

Aleonysh [2.5K]

Answer:

intensity of sound at level of microphone is 0.00139 W / m 2

sound intensity level at position of micro phone is 91.456 dB

EXPLANATION:

Given data:

power of sound P = 31 W

distance betwen microphone & speaker is 42 m

a) intensity of sound at microphone is calculated as

$I = \frac{P}{A}$

$= \frac{34}{4 \pi ( 44m )^ 2}$

= 0.00139 W / m 2

b) sound intensity level at position of micro phone is

$\beta = 10 log \frac{I}{I_o}$

where I_o id reference sound intensity and taken as

$= 1 * 10^{-12} W / m 2$

$\beta = 10 log\frac{0.00139}{10^[-12}}$

= 91.456 dB

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3 years ago

The normal boiling point of cyclohexane is 81.0 oC. What is the vapor pressure of cyclohexane at 81.0 oC?

Aloiza [94]

Answer:

The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

Explanation:

Given that,

Boiling point = 81.0°C

Atmospheric pressure :

Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.

The value of atmospheric pressure is

$P=101325\ Pa$

Vapor pressure :

Vapor pressure is equal to the atmospheric pressure.

Hence, The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

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3 years ago

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3 years ago

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