Answer:
The amount of caffeine left after one half life of 5 hours is 15 oz.
Explanation:
Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.
The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.
So that:
After one half life of 5 hours, the value of caffeine that would be left is;
= 15 oz
The amount of caffeine left after one half life of 5 hours is 15 oz.
Answer:the
8/9 h
Explanation:
Height = 1/2 a T^2 now change to T/3
now height = 1/2 a (T/3)^2 =<u> 1/9</u> 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h
if in series one lightbulb burns out the rest are unable to turn on.
In parallel a single light bulb burns out any other light bulbs are able to work.
Parallel is the best to use during holidays.
The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
