If you just type "<span>What is the chemical formula for mercury(I) nitrate?" into google you get the answer but HG(NO3)2 is the correct one.
sorry no one helped you in time hope you passed anyway</span>
Answer:
I love nice people. Doing anything interesting for Thanksgiving?
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Answer:
Explanation:
cSep 20, 2010
well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.
Anonymous
Sep 20, 2010
First you need to solve for time by using
d=(1/2)(a)(t^2)+(vi)t
1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
t=.45 sec
Then you find the horizontal distance traveled by using
v=d/t
1.3m/s=d/.54sec
d=.585m
Then you need to find the time of player B by using
d=(1/2)(a)(t^2)+(vi)t
1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
t=.61 sec
Finally to find player Bs initial horizontal velocity you use the horizontal equation
v=d/t
v=.585m/.61 sec
so v=.959m/s
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.
Mathematically this expression can be given as

Where
A = Surface area of the Object
Stefan-Boltzmann constant
e = Emissivity
T = Temperature (Kelvin)
Our values are given as





Replacing at our equation and solving to find the temperature 1 we have,




Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C