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:
.409 N
Explanation:
For this to balance, the moments around the fulcrum must sum to zero.
On the left you have .21 ( is that down? I will assume it is)
Counterclockwise moments :
.21 * 40 + 1.0 * 20
Clockwise moments :
.5 * 20 + F * 45
these moments must equal each other
.21*40 + 1 *20 = .5 * 20 + F * 45
F = .409 N
The given mass is 0.025563 g.
Examine the given choices.
a. 0.026 g
This uses 2 significant digits, with rounding to the 3rd decimal place.
b. 2.5 x 10² g = 250 g.
It is incorrect.
c. 0.025 g.
This uses 2 significant digits. It is inaccurate because it does not use rounding to the 3rd decimal place.
d. 0.02 g
This uses one significant digit. It is incorrect for representing the given data.
Answer: a. 0.026 g
Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F
F = 0.679 N
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