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3 years ago

What is the difference between celestial and terrestrial in planetary terms? Someone please answer

1 answer:

5 0

"Celestial" = anything to do with the sky

("Cielo" ..... Spanish for "sky"
"Ceiling" ... that thing up over your head
"Caelum" .. Latin for "heaven")

"Terrestrial" = anything to do with the Earth

("Terra" ... Latin for "Earth")

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4. A steel cable spanning a river is 220.000 m long when the temperature is 30.°C.

romanna [79]

Answer:

219.9208 m

Explanation:

The new length is given by

New length= Original length *(1-Temperature change*coefficient of

thermal expansion of steel)

Here, the change in temperature is $30^{\circ}-0^{\circ}= 30^{\circ}$

New length= $220.000(1-30*12x10^{-6}=219.9208m$

Therefore, the new length will be 219.9208 m

5 0

3 years ago

Which geologic feature would most likely be represented by contour lines drawn far apart from one another?

Anna [14]

It would most definitely not be a hill or a cliff. I believe it would be a plain.

5 0

2 years ago

A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi

Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

5 0

3 years ago

Why are the freezing points of water and the melting point the same

Sladkaya [172]

Because melting point and freezing point describe thesame transition of matter, in this case from liquid to solid (freezing) or equivalently, from solid to liquid (melting).

5 0

3 years ago

Read 2 more answers

34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf

ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as