1. 7.1111m/s Recurring
Solution a = v/t basically divide the velocity by the time taken
2. 9.375m/s
Same solution as before
Question:
A cork floats on the surface of an incompressible liquid in a container exposed to atmospheric pressure. The container is then sealed and the air above the liquid is evacuated. The cork:
A. sinks slightly
B. rises slightly
C. floats at the same height
D. bobs up and down about its old position
Answer:
The correct answer is C) floats at the same height
Explanation:
The liquid is incompressible because its density very high and leaves no room for further compaction whether or not there is atmospheric pressure. So when you put a cork on the liquid, pressure or no pressure, there is no displacement hence it floats on the same height regardless of the absence of air.
Cheers!
First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.
To solve this, you will want to create a proportion.
Flagpole height/flagpole shadow=building height/building shadow
Therefore, it should look like this:
50/30= 300/x
Solve for x:
50x=9,000
X= 180 feet
YOUR ANSWER IS C.
Answer:
Approximately 
(assuming that the projectile was launched at angle of 
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing 
is the same as the altitude 
at which this projectile was launched: 
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 
(upwards,) the vertical velocity right before landing would be 
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be 
. In other words, 
.
Hence, the time it takes to achieve a (vertical) velocity change of 
would be:
.
Hence, this projectile would be in the air for approximately .
This means acceleration a is constant.
Let
a) vo be the initial speed, at t=0
b) v be the final speed after time t
c) d distance travelled in time t
Then we have:
a) v=vo+a×t
b) v²=vo²+2×a×d (Galilei's equation)
c) d=vo×t+a×t²/2
d) average speed vm=(vo+v)/2
