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2 years ago

50 points!

1 answer:

5 0

From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM
M= mass of the Earth = 6×10²⁴ Kg.
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴
T² = 1653339719
∴ T = 40661.28 seconds.

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You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 2.1 m/s. The

garri49 [273]

Answer:

4.6 Joules

Explanation:

K=1/2*MV^2

1/2 * 2.1kg * 2.1^2m/s

==4.6305 Joules

simplified to 4.6 Joules

7 0

2 years ago

In case of collision of objects in two dimensions which statement is true after the collision?

Grace [21]

The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved.

EXPLANATION :

Before coming into any conclusion, first we have to understand the law of conservation of momentum.

As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.

Hence, during any type of collision, the total momentum is always conserved.

Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.

5 0

3 years ago

Read 2 more answers

When is the universal theme of a story often revealed?

elena-14-01-66 [18.8K]

It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>

7 0

3 years ago

Read 2 more answers

Help please<br> It’s kinda urgent

user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

$(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]$

The negative sign of acceleration means that the ship slows down its velocity in order to land.

4 0

2 years ago

A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

$\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}$

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0

1 year ago