Answer:
maximum shear stress = τ(max) = 95.49 × 10⁶N/m²
Explanation:
given
outer diameter at one end(D₁) = 3.0cm
outer diameter at the other end(D₂) = 4.0cm
inner diameter(d₁) = 1.0cm
torque applied(T) = 500Nm
maximum shear stress will occur at lower outer diameter
the formula is τ/r = T/J
τ= T × r/J
where r is radius
T is the torque
J is the polar 2nd M of area
attached is the calculation of the question
a solution with a pH of 2.0 is strongly acidic
Answer:
Vf = 10.76 m/s
Explanation:
Train kinematics
The train moves with uniformly accelerated movement
Formula (1)
Vf: Final speed (m/s)
V₀: Inital speed (m/s)
t: time in seconds (s)
a: acceleration (m/s²)
Movement from t = 0 to t = 5.2s
We replace in formula (1)
4.6 = 0 + a*5.2
a = 4.6/5.2 = 0.88 m/s²
Movement from t = 5.2s to t = 5.2s + 7s = 12.2s
We replace in formula (1)
Vf = 10.76 m/s
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
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
