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dedylja [7]
3 years ago
11

What is answer of question 7

Physics
2 answers:
Stella [2.4K]3 years ago
8 0
Option A it helps us to work
kondaur [170]3 years ago
4 0
A. (They help us work)
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There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

   = 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s

7 0
3 years ago
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An airplane was 300 km [N] of Toronto airport, 3 hours later the airplane 600 km [S]
Alecsey [184]

Answer:

Explanation:

a) 300 + 600 = 900 km S

b) 900/3 = 300 km/hr S

8 0
3 years ago
Which of the following is likely to contribute to geological events that take place on Earth?
Dennis_Churaev [7]
Crust sitting on top of Milton rock of the mantle
5 0
3 years ago
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Why is the sun necessary for life on Earth
Margaret [11]
The sun provides light and energy for plants to live and we need plants to breathe
I hope this helps you
8 0
3 years ago
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You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a
Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

d_x = 5 m

while the vertical distance covered must be

d_y = 6 m

The rock is thrown horizontally with velocity v_x, so we can rewrite the horizontal distance as

d_x = v_x t

where t is the time of flight. Re-arranging the equation,

t=\frac{d_x}{v_x} (1)

The vertical distance covered instead is

d_y = \frac{1}{2}gt^2

where we omit the term ut since the initial vertical velocity is zero. From this equation,

t=\sqrt{\frac{2d_y}{g}} (2)

Equating (1) and (2), we can solve the equation to find v_x:

\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s

6 0
3 years ago
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