Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change
Answer:
This procces is called evaporation.
Explanation:
When you have liquid water that is transformed into steam, a phase change is called evaporation. The temperature for the evaporation of water depends on the pressure, for example for water at atmospheric pressure the temperature of evaporation is equal to 100°C. as the pressure increases are achieved evaporation temperatures higher. When that happens, the phase change temperature of the water is not increasing, as the process that takes place is the transfer of latent heat and applies only to changes of phase, that is to say at atmospheric pressure when it has 100% of the steam this will be at 101°C.
Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation
∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
Answer:
G = 6,786 10⁻¹¹ m³ / s² kg
Explanation:
The law of universal gravitation is
F = G m M/ r²
Where G is the gravitational constant, m and M are the masses of the bodies and r is the distance from their centers
Let's use Newton's second law
F = m a
The acceleration is centripetal
a = 
We replace
G m M / r² = m
G = r² / M
Let's replace and calculate
G = 2.7 10⁻³ (3.88 10⁸)² / 5.99 10²⁴
G = 6,786 10⁻¹¹ m³ / s² kg
Let's perform a dimensional analysis
[N m²/kg²] = [kg m/s² m² / kg²] = [m³ / s² kg]
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
= r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
= r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as
= 0.42 m/s²
