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natali 33 [55]
3 years ago
9

A cylindrical space colony 8km in diameter and 30 km long hasbeen proposed as living quarters for future space explorers. Such a

habitat would have cities, land and lakes on the insidesurface and air and clouds in the center. All this would beheld in place by the rotation of the cylinder about the longaxis. How fast would such a cylinder have to rotate toproduce a 1-g gravitational field at the walls of thecylinder?
a) 0.05 rad/s
b) 0.10 rad/s
c) 0.15 rad/s
d) 0.20 rad/s
e) 0.25 rad./s
Physics
1 answer:
Radda [10]3 years ago
4 0

Answer:

A.

Explanation:

To solve this problem it is necessary to take into account the concepts related to angular velocity and centripetal acceleration,

from the cinematic equations of motion we know that

V= r\omega

where,

V = Tangential Velocity

r = radius

\omega = Angular velocity

While the centripetal acceleration is given by

a = r \omega^2

The displacement of the object is based on gravitational acceleration, so

a = g = 9.8 and the radius would be

r = \frac{d}{2} = \frac{8}{2} = 4000m

Reemplazando:

\omega^2=\frac{g}{R}

\omega= \sqrt{\frac{g}{R}}

\omega = \sqrt{\frac{9.8}{4000}}

\omega = 0.04949rad/s \approx 0.5rad/s

Therefore the answer is A.

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a compound is 24.7% calcium, 1.2% hydrogen, 14.8% carbon, and 59.3% oxygen. write the empirical formula and name the compound.
Tema [17]
For simplicity, assume that the total mass of the compound is 100 g.

Therefore, by mass
Ca = 24.7 g, H = 1.2 g, C = 14.8 g, and O = 59.3 g

Converting these to moles

Mol = mass*1/atomic mass
Ca = 24.7*1/40.078 =0.6163 mol Ca
H = 1.2*1/1.01 = 1.1881 mol H
C = 14.8*1/12.01 = 1.2323 mol C
O = 59.3/16 = 3.7062 mol O

Next, divide all the mols by the smallest value obtained.

Ca: 0.6163/0.6163 = 1 mol Ca
H: 1.1881/0.6163 = 2 mol H
C: 1.2323/0.6163 = 2mol C
O: 3.7062/0.6163 = 6 mol O

Therefore, empirical formula of the compound is

CaH2C2O6

This compound is referred to as Calcium Bicarbonate
8 0
3 years ago
A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the
emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

<u>d = 0.87 m</u>

5 0
3 years ago
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