Answer:
Explanation:
1 ) tire of radius 0.381 m rotating at 12.2 rpm
12.2 rpm = 12.2 /60 rps
n = .20333 rps
angular speed
= 2πn
= 2 x 3.14 x .20333
= 1.277 rad / s
2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s
angular speed = .456 rad/s
3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second
18.7° per second = (18.7 / 180) x 3.14 rad/s
= .326 rad/s
4 )
a rock on a string being swung in a circle of radius 0.587 m with
a centripetal acceleration of 4.53 m/s2
centripetal acceleration = ω²R
ω is angular velocity and R is radius
4.53 = ω² x .587
ω = 2.78 rad / s
5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
The radius of the circle in which corner is moving
= .123 x √2
=.174 m
angular velocity = linear velocity / radius
.287 / .174
1.649 rad / s
The perfect order is
4 ) > 5> 1 >2>3.
B. Australia and Antarctica were a single landmass in the past.
Explanation:
The most likely explanation for why the fossils of this plant are found on these two continents is that Australia and Antarctica were a single landmass in the past.
This evidence is vastly supported by the theory of plate tectonics originally, the theory of continental drift.
- Plate tectonics proposes that all land masses were once joined together as a single supercontinent called Pangaea.
- The region of Antartica and Australia were once joined as a single continental mass.
- The plant must have developed together at that time.
- When the continents moved apart, they still had the fossils.
learn more:
Continental drift brainly.com/question/5002949
#learnwithBrainly
The 'Bulge', the 'Disk', the 'Spiral Arms', and the 'Halo' those are four parts of a spiral galaxy. Try finding out the many different types of galaxies to understand which is which.
Answer:
a) 14.2 atm
b) 4.46 atm
c) 1.06 atm
Explanation:
For an ideal gas,
PV = nRT
P = pressure of the gas
V = volume occupied by the gas
n = number of moles of the gas
R = molar gas constant = 0.08206 L.atm/mol.K
T = temperature of the gas in Kelvin
a) For HF,
P =?, V = 2.5L, n = 1.35 moles, T = 320K
P = 1.35 × 0.08206 × 320/2.5
P = 14.2 atm
b) For NO₂
P =?, V = 4.75L, n = 0.86 moles, T = 300K
P = 0.86 × 0.08206 × 300/4.75
P = 4.46 atm
c) For CO₂
P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K
P = 2.15 × 0.08206 × 330/55
P = 1.06 atm
