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Blababa [14]
3 years ago
9

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s^2 at the rim?
Physics
1 answer:
Anestetic [448]3 years ago
5 0

Answer:

ω = 0.313 rad/s

Explanation:

D = 200 m ⇒  R = D / 2 = (200 m / 2) = 100 m

ac = 9.8 m/s²

The following equation can be used

ac = ω²*R    ⇒    ω = √(ac / R)

then we insert the values in the last formula

ω = √(9.8 m/s² / 100 m) = 0.313 rad/s

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Balancing Chemical Equations
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1.   2 C2H6 + 7 O2 = 6 H2O + 4 CO2

8 0
3 years ago
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A motorcycle has a constant acceleration of 3.49 m/s2. Both the velocity and acceleration of the motorcycle point in the same di
Vilka [71]

Answer:

(a)2.865 s

(b)2.865 s

Explanation:

We are given that

Acceleration,a=3.49 m/s^2

a.Initial speed,u=29 m/s

Final speed,v=39 m/s

We know that

t=\frac{v-u}{a}

Using the formula

t=\frac{39-29}{3.49}=2.865 s

b.Initial speed,u=59 m/s

Final speed,v=69 m/s

Again using the formula

t=\frac{69-59}{3.49}=2.865 s

7 0
3 years ago
How much does a 0.15 kg baseball weigh on earth?
Gnesinka [82]
1.472 N

to get weight you multiply an object's mass in kilograms with the acceleration of gravity(9.81m/s) :)
7 0
3 years ago
A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0
sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball  is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})

Since the initial height is 0

Solving for v_{o}

v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s

5 0
3 years ago
Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a
dusya [7]

Answer:

W=-21,870,000\ J

Explanation:

<u>Work and Kinetic Energy </u>

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

W=\Delta k=k_1-k_0

Since

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}

W=0-21870000\ J

\boxed{W=-21,870,000\ J}

3 0
3 years ago
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