True; scientist define work as the spending of energy. This is because energy is required to move an object. For work to be done an object has to be moved from one point to another. Work is measured in joules similar to energy which is also measured in joules. Mathematically, work done is given by the product of force and distance moved by an object or a body.
I think it would be water but im not sure! Hope that helps!!
Answer:
The amount of drug left in his body at 7:00 pm is 315.7 mg.
Explanation:
First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:
Where:
λ: is the decay constant = 
: is the half-life of the drug = 3.5 h
N(t): is the quantity of the drug at time t
N₀: is the initial quantity
After 90 min and before he takes the other 200 mg pill, we have:
Now, at 7:00 pm we have:
Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).
I hope it helps you!
Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
