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3 years ago

Is the moon flat tell me if it it

Physics

2 answers:

QveST [7]3 years ago

6 0

I don’t think it is flat

yan [13]3 years ago

6 0

Answer: The moon is not flat.

Explanation: To the eye, the moon appears round, and it's natural to assume that it is actually spherical in shape with every point on its surface equidistant from its center like a big ball. The shape of the moon is that of an oblate spheroid, meaning it has the shape of a ball that is slightly flattened.

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What is the pressure in a gas container that is connected to an open-end U-tube manometer if thepressure of the atmosphere is 75

VARVARA [1.3K]

Answer:

The pressure in a gas container is 839.44mmHg

Explanation:

Density of Mercury = 13.56g/cm³ = 13556kg/m³

$P_{A} = P_{0} +$ hρg/ P₀

where, g= acceleration due to gravity, 10m/s²

h= height= 8.6cm = 0.086m

$P_{A}$ = hρg/ P₀

= 0.086 × 13556 × 10

= 11658.16 kgm/ m²s²

= 11658.16 N/m² = 11658.16pa

= 87.4434mmHg

where 1 torr = 1mmHg

$P_{0}$ = 752torr = 752mmHg

$P_{A$ = 752 + 87.4434

= 839.44mmHg

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3 years ago

A less-dense liquid of density rho1 floats on top of a more-dense liquid of density rho2. A uniform cylinder of length l and den

Umnica [9.8K]

Answer:

Explanation:

Given

$\rho=density\ of\ cylinder\\\rho_1=less\ dense\ cylinder\\\rho_2=more\ dense\ cylinder$

Suppose V is the volume of a cylinder

Also $L_1=length\ in\ less\ dense\ part\\L_2=length\ in\ dense\ part\\L=length\ of\ cylinder\\V=A\times L\\where\ A=area\ of\ cross-section\\$

Now, we can write

The weight of the cylinder is supported by the buoyant forces of two liquids

$\rho Vg=\rho_1 V_1g+\rho_2 V_2g\\\rho ALg=\rho_1 A\times L_1g+\rho_2 A\times L_2g\\\\\rho L=\rho_1 L_1+\rho_2 L_2\\Also, L=L_1+L_2$

using the above equation we can write

$L_2=\frac{\rho-\rho_1}{\rho_2-\rho_1}\cdot L\\\frac{L_2}{L}=\frac{\rho-\rho_1}{\rho_2-\rho_1}$

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3 years ago

What type of fault usually occurs because of compression?

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The answer is:
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3 years ago

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3 years ago

Read 2 more answers

You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago