Answer:
The atmospheric pressure and boiling point are directly proportional
Increasing atmospheric pressure increases the boiling point also
Explanation:
The atmosphere contain molecules that are in constant motion. They exert a downward force on a liquid’s surface. The higher the air pressure, the harder it is for the liquid to evaporate. Therefore, the boiling point of a solvent or liquid is affected by the atmospheric pressure and boiling point is raised.
A liquid in a high pressure environment boils at a higher temperature.
When placed in a lower pressure environment it boils at a lower temperature.
A. Impulse is simply the product of Force and time.
Therefore,
I = F * t --->
1
where I is impulse, F is force, t is time
However another formula for solving impulse is:
I = m vf – m vi --->
2
where m is mass, vf is final velocity and vi is initial
velocity
Therefore using equation 2 to solve for impulse I:
I = 2000kg (0) – 2000kg (77 m/s)
I = -154,000 kg m/s
B. By conservation of momentum, we also know that Impulse
is conserved. That means that increasing the time by a factor of 3 would still
result in an impuse of -154,000 kg m/s. So,
I = F’ * (3 t) = -154,000 kg m/s
Since t is multiplied by 3, therefore this only means
that Force is decreased by a factor of 3 to keep the impulse constant,
therefore:
(F/3) (3t) = -154,000 kg m/s
Summary of Answers:
A. I = -154,000 kg m/s
B. Force is decreased by factor of 3
Answer:
it should be B let me know if i am wrong
Explanation:
Answer:
-5.9 rad/s^{2}
Explanation:
radius (r) = 30 cm = 0.3 m
mass (m) = 1.05 kg
initial speed (u) = 77 rpm
final speed (v) = 0 rpm
time (t) = 1.37 s
angular acceleration =
therefore
initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s
final speed (v) = 0 rpm = 0 rad/s
angular acceleration = 
= -5.9 rad/s^{2}
The answer to this question is TRUE.
