Answer:
1. 1 s = 1 x 10⁶ μs
2. 1 g = 0.001 kg
3. 1 km = 1000 m
4. 1 mm = 1 x 10⁻³ m
5. 1 mL = 1 x 10⁻³ L
6. 1 g = 100 dg
7. 1 cm = 1 x 10⁻² m
8. 1 ms = 1 x 10⁻³ s
Explanation:
1.
1 x 10⁻⁶ s = 1 μs
(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs
<u>1 s = 1 x 10⁶ μs</u>
2.
1000 g = 1 kg
1 g = 1/1000 kg
<u>1 g = 0.001 kg</u>
3.
<u>1 km = 1000 m</u>
<u></u>
4.
<u>1 mm = 1 x 10⁻³ m</u>
<u></u>
5.
<u>1 mL = 1 x 10⁻³ L</u>
<u></u>
6.
1 x 10⁻² g = 1 dg
(1 x 10⁻² x 10²) g = 1 x 10² dg
<u>1 g = 100 dg</u>
<u></u>
7.
<u>1 cm = 1 x 10⁻² m</u>
<u></u>
8.
<u>1 ms = 1 x 10⁻³ s</u>
False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.
Answer:
3.57 MJ
Explanation:
ASSUMING it's fresh water with density of 1000 kg/m³
W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J
Salt water would require more.
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
