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3 years ago

Which statement best describes perigee?

1 answer:

algol133 years ago

7 0

<span>
A. The closet point in the Moon's orbit to Earth . . . . . perigee

B. The farthest point in the Moon's orbit to Earth . . . . . apogee

C. The Sun's orbit that is closest to the Moon . . . . . a meaningless description

D. The closest point in Earth's orbit of the Sun . . . . . perihelion

-- The farthest point in Earth's orbit of the Sun . . . . . aphelion
</span>

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A spider of mass mm is swinging back and forth at the end of a strand of silk of length LL. During the spider's swing the strand

krok68 [10]

Answer:

The speed of the spider is v = (2g*L*(1-cosθ))^1/2

Explanation:

using the energy conservation equation we have to:

Ek1 + Ep1 = Ek2 + Ep2

where

Ek1 = kinetic energy = 0

Ep1 = potential energy = m*g*L*cosθ

Ek2 = (m*v^2)/2

Ep2 = m*g*L

Replacing, we have:

0 - m*g*L*cosθ = (m*v^2)/2 - m*g*L

(m*v^2)/2 = m*g*L*(1-cosθ)

v^2 = 2g*L*(1-cosθ)

v = (2g*L*(1-cosθ))^1/2

4 0

3 years ago

What are the answers according to Henry's classification.

Novosadov [1.4K]

Answer:

Hope this helps you!

5 0

2 years ago

The sputnik one satellite orbiting earth mass equals 5.98×10 to the 24th power kilograms in a circle of radius 6.96×10 to the si

nika2105 [10]

Answer:

7572 m/s

Explanation:

The force between two masses separated by a distance r is given as:

$F=G\frac{m_1m_2}{r^2}$

Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is $6.674*10^-11 Nm^2/kg^2$

The mass of the earth ( $m_1$ ) is far greater than the mass of the sputnik ( $m_2$ ). Therefore $m_1m_2=m_1$ . The mass of the sputnik is neglected, therefore:

$F=G\frac{m_1}{r^2}=\frac{(6.674*10^{-11})(5.98*10^{24})}{(6.96*10^6)^2} = 8.2389N$

But F is actually centripetal acceleration, a = v²/r

$8.2389 = v^2 / 6.96*10^6\\v=7572m/s$

6 0

2 years ago

A stone is dropped from a certain height, distance covered by it in one second is:

Alinara [238K]

Answer:

d = 4.9 m

Explanation:

It is mentioned that a stone is dropped from a certain height. It is required to find the distance covered by it in one second.

The initial speed of the stone is equal to 0 as it was at rest. Let d is the distance covered by the stone.

Using second equation of motion :

$d=ut+\dfrac{1}{2}at^2$

Put u = 0 and a = g

$d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times 1^2\\\\d=4.9\ m$

So, the distance covered by it in one second is 4.9 m.

5 0

3 years ago

The fragment of an asteroid or any interplanetary material is known as a: A) limestone dignitary satellite. B) moon. C) shower m

ElenaW [278]

The fragment of an asteroid or any interplanetary material is known as a a : D. Meteroid

Human came in contact with this material mostly because it penetrate the atmosphere and fall within the earth surface

hope this helps

6 0

2 years ago