Answer:
You are pulled towards that building. At the same time, that building is pulled towards you. Neither object creates enough gravitational force to really do anything. That is why you never notice any affect by either body, (you and a building).
Explanation:
You will surely get attracted towards the building.But it takes a lot of time depending on their masses.
This happens only when you are away from earth with that building.
Both of you will get attracted to it
if a third party with mass more than you or building is with you.
If it is on the earth.. Then the gravity between you and the building is negligible compared to the earth.Hence you will not get attracted towards the building in this case.
We see more and more of the lighted side of the moon.
you're welcome, m8
Answer:
So the sound intensity level they would experience without the earplugs is 110.32dB.
Explanation:
Given data
Sound intensity by factor =215
Sound intensity level =87 dB
To find
Sound intensity level they would experience without the earplugs
Solution
First we need to find the new sound intensity level
So

The dB can be calculated as:

Substitute the given values

So the sound intensity level they would experience without the earplugs is 110.32dB.
its B. 60 meters
Explanation:
cause I looked up a calculator and solved it
Answer:

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

Explanation:
For this case we know the mass of the water given :

And we know that the initial temperature for this water is
.
We want to cool this water to the human body temperature 
Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Where
represent the specific heat for the water and this value from tables we know that
for the water.
So then we have everything in order to replace into the formula of sensible heat and we got:

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:
