Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
Answer:
Δ v = 125 m/s
Explanation:
given,
mass of space craft = 435 Kg
thrust = 0.09 N
time = 1 week
= 7 x 24 x 60 x 60 s
change in speed of craft = ?
Assuming no external force is exerted on the space craft
now,
a = 2.068 x 10⁻⁴ m/s²
using equation of motion
Δ v = a t
Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60
Δ v = 125 m/s
Answer:
π/10 rads
Explanation:
It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of
2πl rad.
Now, number of periods of 3 minutes in an hour is;
Number of periods = 60/3 = 20 periods
Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.
Thus;
Angular displacement = (1/20) * 2π = π/10 rads
