A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.
1 answer:
Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 = × 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
You might be interested in
Answer:
4.156 m/s
Explanation:
See attachment
Answer:
Time, t = 0.87 seconds
Explanation:
Given that,
Initial velocity of the object, u = 4.3 m/s
The coefficient of kinetic friction between horizontal tabletop and the object is 0.5
We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.
Using first equation of motion to find it as :
a is the acceleration, here,
So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
Answer:
The acceleration is in 2 D as in between east and south.
Explanation:
mass, m = 50 kg
acceleration, a = 0.25 m/s^2 horizontal
acceleration of elevator, a' = 1 m/s^2 downwards
When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:
