Answer:
Hydrochloric and Hydrofluoric Acids.
Answer:
There is an interval of 24.28s in which the rocket is above the ground.
Explanation:





From Kinematics, the position
as a function of time when the engine still works will be:

At what time the altitud will be
?
⇒ 
Using the quadratic formula:
.
How much time does it take for the rocket to touch the ground? No the function of position is:

Where our new initial position is
, the velocity when the engine breaks is
and the only acceleration comes from gravity (which points down).
Now, when the rocket tounches the ground:
Again, using the quadratic ecuation:

Now, the total time from the moment it takes off and the moment it tounches the ground will be:
.
There is half the force that there was before it was split in half
<h3>
Answer: 22.5 m/s</h3>
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Work Shown:
acceleration = ( finalVelocity - initialVelocity )/(change in time)
1.5 = (60 - x)/(25)
1.5*25 = 60-x
37.5 = 60-x
x = 60-37.5
x = 22.5
The initial velocity is 22.5 m/s
Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
= 30 J, l = 10 cm = 0.1 m (as 1 m = 100 cm)
F = 10 N
The formula to calculate the work done is as follows.
W = 
Hence, putting the given values into the above formula as follows.
W = 
= 
= 1 J
As the dart loses 1 joule of energy when it goes through the barrel. Therefore, the kinetic energy it has when it comes out is calculated as follows.
K.E = 30 - 1
= 29 J
Thus, we can conclude that the kinetic energy of the fired dart is 29 J.