Liquids is transformed feom a liquid state into a gaseous state. this can happen at any temperature
Answer:
α = 0
, w = w₀
Explanation:
Torque is related to angular acceleration by Newton's second law for rotational motion.
τ = I α
Where τ is the torque, I the moment of inertia and α the angular acceleration.
If we apply an external torque for the sum of all torques to be zero, the angular acceleration must fall to zero
α = 0
Since the acceleration is zero, the angular velocity you have at that time is constantly killed.
w = w₀ + α t
w = w₀ + 0
Answer: 0.790 g/cm3
Explanation:
The density of acetone is 790 Kg/m3.
To convert from Kg to g we multiply by 1000 (1 Kg = 1000 g)
To convert from m3 to cm3 we multiply by 10∧6
So, The density of acetone in (g/cm3) = (790 x 1000) / (10∧6) = 0.79 g/cm3
The difference is (80 dB - 40 dB) = 40 dB.
The sound is 40 dB louder outside.
Each 10 dB means 10 times more sound power.
40 dB louder means 10x10x10x10 times more sound power.
That's <em>10,000 times</em> more sound power outside than inside.
Complete question:
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.
(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Answer:
The net force on the person as the air bad deploys is -6750 N backwards
Explanation:
Given;
mass of the passenger, m = 60 kg
velocity of the car at impact, u = 15 m/s
final velocity of the car after impact, v = 0
distance moved as the front of the car crumples, s = 1 m
First, calculate the acceleration of the car at impact;
v² = u² + 2as
0² = 15² + (2 x 1)a
0 = 225 + 2a
2a = -225
a = -225 / 2
a = -112.5 m/s²
The net force on the person;
F = ma
F = 60 (-112.5)
F = -6750 N backwards
Therefore, the net force on the person as the air bad deploys is -6750 N backwards
