<span>One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle, theta, with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.390. A. what is the maximum value...</span>
if we are walking on a perfectly smooth ground which has no friction our force would simply cancel out the force reverted by the ground and we would fall.
We need it to help push out feet off the ground
Hope those helps :)
Why this would occur is because when the ground shakes under the water it causes the earth to move, which means earths plates are shaking which can lead to what?
Answer:
τ = (7.96 x 10⁴ m⁻³)T
This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.
Explanation:
The maximum allowable shear stress on the solid shaft can be given by the torsional formula as follows:
τ = Tc/J
where,
τ = Maximum Allowable Shear Stress = ?
T = Maximum Torque Applied to the Shaft
c = maximum distance from center to edge = radius in this case = 20 mm = 0.02 m
J = Polar Moment of inertia = πr⁴/2 = π(0.02 m)⁴/2 = 2.51 x 10⁻⁷ m⁴
Therefore,
τ = T(0.02 m)/(2.51 x 10⁻⁷ m⁴)
<u>τ = (7.96 x 10⁴ m⁻³)T</u>
<u>This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.</u>
Answer:
the buoyant force on the chamber is F = 7000460 N
Explanation:
the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.
Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume
mass of water displaced = density of seawater * volume displaced
m= d * V , V = 4/3π* Rext³
the buoyant force is the weight of this volume of seawater
F = m * g = d * 4/3π* Rext³ * g
replacing values
F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N
Note:
when occupied the tension force on the cable is
T = F buoyant - F weight of chamber = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N
