I dont think so since the alcohol is still mixed into the water so I think they will evaporate at the same time.
1). If you want the wheel to turn as one solid piece, then all parts of it must have the same RATE of rotation. If one part was going 2 RPM and another part was going 3 RPM, then there's no way both parts could stay hooked together. Their SPEED depends on their distance from the center, but they all have to make the same RPM. 2). The CENTRIPETAL acceleration of anything that's rotating is. m-v-squared/ R. The object's speed depends on its radius, and it's acceleration varies directly with the square of the speed. So if you move in to half the radius, the acceleration becomes 1/4 of the original value.
Answer:
The time the train blocked the crossway is
0.007934h=0.476 min=28.6 s
Explanation:
The average speed is given by
V= (Vi+Vf)/2
and
V= dx/dt
Where
Vi= initial speed
Vf= final speed
dx= change of position=0.384 km
dt=time the train blocked the crossway
dt=2dx/(Vi+Vf)
dt=2*0.384 km/(77.3 +19.5) km/h
dt=0.007934h
dt=0.007934h*(60min/1h)=0.476 min
dt=0.476 min*(60s/1min)=28.6 s
Answer:
the engine cool to 40
at 14.07 minutes
Explanation:
Given information
T(5) = 70
= 100
C = 15
Newton's law of cooling :
T(t) = C + ( - C) 
where
T(t) = temperature at any given time
C = surrounding temperature
= initial temperature of heated object
k = cooling constant
to find the the time when the engine will be cooled down to 40, we first need to find the cooling constant, k
when t = 5, T(5) = 70
so,
T(t) = C + ( - C)
T(5) = 15 + (100 - 15) 
70 = 15 + (85)
= (70 - 15) / 85
-5k = ln (55/85)
k = - ln (55/85) / 5
k = 0.087
thus, we have the eqaution
T(t) = 15 + (85) 
now we can determine the time when T(t) = 40
40 = 15 + (85)
= (40-15)/85
-0.0087t = ln (25/85)
t = - ln (25/85)/0.087
t = 14.07 minutes
The distance traveled while accelerating from rest is
D = 1/2 a t² .
For this problem, we shall totally ignore air resistance.
We do so completely without any reservation or guilt,
because we know that there is no air on the moon.
D = (1/2) · (1.6 m/s²) · (9 sec)²
= (0.8 m/s²) · (81 s²)
= (0.8 · 81) m
= 64.8 meters .
(That's about 213 feet ! The astronaut must have dropped the feather
from his spacecraft while he was aloft ... either just before touchdown
or just after liftoff.)
