Answer: 0.98m
Explanation:
P = -74 mm Hg = 9605 Pa = 9709N/m^2
= 9605 kg m/s^2/m^2
density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3
Pressure equation: P = rho g h
h = P/(rho g)
h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)
h = 0.98 m
0.98m is the maximum depth he could have been.
Answer:
F₁ = F₂ = F₃ = 0 N
Explanation:
given,
Arrow 1 mass = 80 g speed = 10 m/s
Arrow 2 mass = 80 g speed = 9 m/s
Arrow 3 mass = 90 g speed = 9 m/s
Horizontal Force:- F₁ , F₂ and F₃
There is no air resistance.
If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.
We know,
According to newton's second law
F = m a
If Acceleration is equal to zero
Then Force is also equal to zero.
Hence, F₁ = F₂ = F₃ = 0 N
Answer:
the rms speed of cesium atoms that have been cooled to a temperature of 100nK = 0.43cm/s or 0.0043m/s
Explanation:
The concept of root mean square velocity is applied, where the average translational kinetic is related to the actual kinetic energy, the expression for the root mean square is the generated.
The detailed steps and appropriate substitution is as shown in the attachment.
Answer:
F=126339.5N
Explanation:
to find the necessary force to escape we must make a free-body diagram on the hatch, taking into account that we will match the forces that go down with those that go up, taking into account the above we propose the following equation,
Fw=W+Fi+F
where
Fw= force or weight produced by the water column above the submarine.
to fint Fw we can use the following ecuation
Fw=h. γ. A
h=distance
γ=
specific weight for seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w is the weight of the hatch = 200N
Fi is the internal force of the submarine produced by the pressure = 1atm = 101325Pa for this we can use the following formula
Fi=PA=101325x0.7=70927.5N
finally the force that is needed to open the hatch is given by the initial equation
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
