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3 years ago

Which statement supports the giant-impact hypothesis of the moon’s formation

Physics

2 answers:

V125BC [204]3 years ago

7 0

Answer:

the moon lacks a sizeable iron core

Explanation:

Katen [24]3 years ago

5 0

Answer:

The moon lacks a sizable iron core is the answer

Explanation:

<h2>Got it correct on BRAINLY <----≡</h2>

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Please help (will mark brainliest)

serg [7]

Answer:

if im not mistaken i think its d let me know if correct plz

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2 years ago

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A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

Nady [450]

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

$v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s$

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

8 0

3 years ago

Seja uma carga Q=1.2×10(-8)C no vácuo, distando 40 cm de um ponto M e 50 cm de um ponto N . Determine os pontencias no ponto M e

Len [333]

<span>Americano, por favor?</span>

4 0

3 years ago

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

$\rho = 12580.7 kg/m^3$

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3 years ago

Which of Newton’s Laws involves mass and acceleration? Question 1 options: 3rd 1st 2nd All of them

Feliz [49]

The answer would be Newton’s Second Law

6 0

3 years ago

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