Depth = 5.0 × 10^2 m
Density of seawater = 1.025 x 10^3
Pd = Po + pgh
Standard atmospheric pressure is Patm = 1.01325 x 10^5 Pa
Since the pressure inside the hull is normal, we can disregard Po.
Thus, Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
Now calculating Pd / Patm gives us 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
This indicates the pressure is 49.56 times greater.
Answer:
Explanation:
at 23 degrees Celsius, the diameter measures 4.511 mm
GIVEN DATA:
diameter of hole = 4.500 mm
T_1 = 23.0 degrees Celsius
T_2 = - 78.0 degrees Celsius
the expansion coefficient of aluminum is 2.4*10^{-5} (degrees Celsius)^{-1}
the diameter at 23 degrees Celsius is stated as
= 4.511 mm
the diameter of the rivet after temperature change is given as
= 0.4511 cm
Referencing the diagram below, we can deduce from the geometry that x = 2.5 - 0.55 = 1.95 m, leading to cos θ = 1.95/2.5 = 0.78. Therefore, θ = cos⁻¹ 0.78 = 38.74°. According to the free body diagram, the tension in the chain measures 450 N. Here, F denotes the centripetal force and W signifies Dee's weight. The tension's components are as follows: Horizontal component = 450 sin(38.74°) = 281.6 N, directed to the left, and Vertical component = 450 cos(38.74°) = 351.0 N, directed upward. Answers: Horizontal: 281.6, directed left. Vertical: 351.0 N, directed upward.
Utilizing the equation F = ma, where F represents the force applied by the machine, A denotes acceleration (equivalent to v/t, with v as velocity and t as time), and M symbolizes mass, we can calculate as follows: F = mv/t. Thus, F = (0.15kg) (30 – 0 m/s) / 0.5 s, resulting in F = 9 N.
Answer:
Distance: 4.6 km Displacement= -0.2 km
Explanation:
The overall distance covered: 1.5 + 2.4 + 0.7 = 4.6 km
Displacement calculation: 1.5 - 2.4 + 0.7 = -0.2 km
The displacement could also simply be stated as 0.2 km depending on whether negative value is preferred.
