Answer:
Explanation:
Given
diameter of spacecraft 
radius 
Force of gravity 
=mg
where m =mass of object
g=acceleration due to gravity on earth
Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by
Answer: 24.97 kg
Explanation:
The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:
F = G*(M1*M2)/R^2
Where G is the gravitational constant:
G = 6.67*10^-11 m^3/(kg*s^2)
In this case, we know that
R = 0.002m
F = 0.0104 N
and that M1 = M2 = M
And we want to find the value of M, then we can replace those values in the equation to get
0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2
(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2
623.69 kg^2 = M^2
√(623.69 kg^2) = M = 24.97 kg
This means that the mass of each object is 24.97 kg
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K
A = 4\pi r^2
A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2
A = 1.33*!0^{-5}MM^2
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