Answer:
Gravitational force (pulled downward by the Earth)
Normal force (pushed upward by the ground)
Applied force (pushed by the person)
Friction force (pulled opposite the direction of motion by the roughness of the ground)
Answer:
σ = 391.2 MPa
Explanation:
The relation between true stress and true strain is given as:
σ = k εⁿ
where,
σ = true stress = 365 MPa
k = constant
ε = true strain = Change in Length/Original Length
ε = (61.8 - 54.8)/54.8 = 0.128
n = strain hardening exponent = 0.2
Therefore,
365 MPa = K (0.128)^0.2
K = 365 MPa/(0.128)^0.2
k = 550.62 MPa
Now, we have the following data:
σ = true stress = ?
k = constant = 550.62 MPa
ε = true strain = Change in Length/Original Length
ε = (64.7 - 54.8)/54.8 = 0.181
n = strain hardening exponent = 0.2
Therefore,
σ = (550.62 MPa)(0.181)^0.2
<u>σ = 391.2 MPa</u>
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top 
= 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
= √2gh
we substitute
= √(2 × 9.81 m/s² × 0.2 m )
= 1.98091 m/s
= 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π
/4) ×
= √(4Q/π)
we substitute our values into the equation;
= √(4(400 cm³/s) / (π×198.091 cm/s))
= 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
B) the plates are in constant motion and as a result the boundaries are where they interact
