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3 years ago

Classify the planets as inner planets of outer planets explain your answer

Physics

1 answer:

rusak2 [61]3 years ago

7 0

Answer: Planets A and B have a rocky mantle and an iron core. They are at a distance of 1 AU or less from the Sun. This means they are relatively close to the Sun. There are the properties of inner (terrestrial) planets. So planet A and B are inner planets.

Explanation:

Planet C is composed if the gases hydrogen and helium. It has a high mass and is much farther from 1 AU from the sun. These properties are consistent with the outer planets. So, planet C is an outer (gas giant) planet.

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Why are we buried in the ground when we die?

trasher [3.6K]

What kind of the question is that. We aren't really buried. We either go to Heaven or the other place. Some people say it's over when your buried in the ground but believers don't really think that.

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3 years ago

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It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet

Amiraneli [1.4K]

Answer:

354.72 m/s

Explanation:

$m$ = mass of lead bullet

$c$ = specific heat of lead = 128 J/(kg °C)

$L$ = Latent heat of fusion of lead = 24500 J/kg

$T_{i}$ = initial temperature = 27.4 °C

$T_{f}$ = final temperature = melting point of lead = 327.5 °C

$v$ = Speed of lead bullet

Using conservation of energy

Kinetic energy of bullet = Heat required for change of temperature + Heat of melting

$(0.5) m v^{2} = m c (T_{f} - T_{i}) + m L\\(0.5) v^{2} = c (T_{f} - T_{i}) + L\\(0.5) v^{2} = (128) (327.5 - 27.4) + 24500\\(0.5) v^{2} = 62912.8\\v = 354.72 ms^{-1}$

3 0

3 years ago

What is the difference between an object’s speed and its acceleration? NO LINKS

Arisa [49]

Answer: Hello! An objects speed is constant and has the units meters per second (m/s); thus, it does not change overtime. Acceleration is a rate of change where the speed does either increase or decrease overtime from its inital value; its units are meters per second second (m/s/s). I hope that helps!

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3 years ago

A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizont

aev [14]

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

$F_y = F sin \theta$

$F_y = 245\times sin 55^0$

$F_y =200.69$

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

6 0

3 years ago

Read 2 more answers

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

3 years ago

Classify the planets as inner planets of outer planets explain your answer​

Classify the planets as inner planets of outer planets explain your answer