<span>1.5 minutes per rotation.
The formula for centripetal force is
A = v^2/r
where
A = acceleration
v = velocity
r = radius
So let's substitute the known values and solve for v. So
F = v^2/r
0.98 m/s^2 = v^2/200 m
196 m^2/s^2 = v^2
14 m/s = v
So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin.
Its circumference is 2*pi*r, so
C = 2 * 3.14159 * 200 m
C = 1256.636 m
And we need a velocity of 14 m/s, so
1256.636 m / 14 m/s = 89.75971429 s
Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.</span>
I assume here that the engine operates following a Carnot cycle, which achieves the maximum possible efficiency.
Under this assumption, the efficiency of the engine (so, the efficiency of the Carnot cycle) is given by
where
is the cold temperature
is the hot temperature
For the engine in our problem, the cold temperature is 313 K while the hot temperature is 425 K, so the effiency of the engine is
Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Answer: Option D : is traveling rapidly but oscillating slowly.
Explanation:
ycarrier(x,t) is traveling rapidly but oscillating slowly.
