What material composes most of the volume of planets jupiter and saturn?

Write Newton's 3rd law. How would this law relate to a rocket ship taking off from the earth? Would this law affect the rocket s

irina1246 [14]

Answer:

Part A

Newton's 3rd law states that action and reaction are equal and opposite, mathematically, we have;

$F_A$ = - $F_B$

Where;

$F_A$ = The action force

$F_B$ = The reaction force

Part B

The law indicates that the force with which a rocket ship uses in taking off from the Earth, $F_A$ is equal in magnitude, and opposite in direction to the reaction force of the Earth to the motion of the rocket, (-) $F_B$

Part C

The law is a universal law, and it will also affect the rocket ship in space, as the force of the jet from the exhaust is directed towards Earth while in space, the rocket is propelled deeper into space

Explanation:

6 0

2 years ago

What's ur fav song? Fr33 Pointtsss <3

arsen [322]

Answer:

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Haha Rick rolled you

Explanation:

jk my favourite song is Thunder, Despacito

3 0

2 years ago

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A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *

SSSSS [86.1K]

Answer:

0.013N

Explanation:

$F=\frac{mv^{2} }{r}$

m=0.04, v=4.4m/s, r=60

F=0.013 N

4 0

2 years ago

During baseball practice, a batter hits a high flyball, and then runs in a straight line and catches it. which has the greater a

Jlenok [28]

They have the same velocity because their displacements (shortest line from point A to point B, which is a straight line) are the same and they meet at the same time.

6 0

3 years ago

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