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Bad White [126]

3 years ago

9

The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t

hat is 2R from the center of the moon. What is the gravitational field strength of the moon at this new distance?

1 answer:

3241004551 [841]3 years ago

6 0

Answer:

$g'=\frac{g__R}{4}$

Explanation:

Given:

gravitational field strength of moon at a distance R from its center, $g__R$

Distance of the satellite from the center of the moon, $h=2R$

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

$g'=g__{R}} \times \frac{R^2}{(2R)^2}$

$g'=\frac{g__R}{4}$

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

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Why is plate tectonics a widely accepted theory?

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Answer: The theory of Plate Tectonics is now widely accepted because there is sufficient proof to support it, and it is an important aspect of geology, oceanography, geophysics and even paleontology.

Explanation: In places where a plate faced resistance to its movement, it would fold upward and create mountains. Hope this helped! :)

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3 years ago

When considering the gas laws, the kelvin temperature scale must be used. the reason for this is that the kelvin scale is direct

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4 years ago

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Ganezh [65]

Answer:

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340/(4 × 0.35) = 242.85 Hz

5 0

3 years ago

Choose the best scientific reasoning for how objects with different amounts of mass can influence the rate at which they slow do

aev [14]

Answer:

The heavier something is, the harder it is to slow down. As such, higher amounts of mass result in a lower rate of slowing.

6 0

3 years ago

What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an

astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution= $T_b=101^{\circ}$ C

Boiling point water=100 degree Celsius

$K_f=1.86K/m$

$K_b=0.512 K/m$

$\Delta T_b=T-T_0$

Where $T$ =Boiling point of solution

$T_0=$ Boiling point of pure solvent

$\Delta T_b=101-100=1^{\circ}$ C

$\Delta T_b=k_bm$

Using the formula

$1=0.512\times m$

Molality, $m=\frac{1}{0.512}$ m

$\Delta T_f=k_fm$

Using the formula

$\Delta T_f=\frac{1}{0.512}\times 1.86$

$\Delta T_f=3.63 C$

We know that

$\Delta T_f=T_0-T_1$

Where $T_0$ =Freezing point of solvent

$T_1=$ Freezing point of solution

Using the formula

$3.63=0-T_1$

Freezing point of water=0 degree Celsius

$T_1=0-3.63=-3.63 C$

Hence, the freezing point of solution=-3.63 degree Celsius

7 0

3 years ago