an astronaut on the Moon could experience zero gravity, and would see stars and Earth.
The resistance of the lamp is apparently 50V/2A = 25 ohms.
When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.
In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.
The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.
The power delivered by the 200V mains is (200V) x (2A) = 400 watts.
The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.
The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.
Together, they add up to the 400 watts delivered by the mains.
CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
I think that the best one is determining one's sense of individuality and place in society.
To contrast inner and outer planets we will start with the climate of the planets and then move on to there lighting. To start the planets closet to the sun, mercury, venus, earth and mars, are all hot compared to the further one, jupiter, saturn, uranus, neptune. This distance also makes the farthe away planets darker than the ones closer. Now to compare all the planets vary from either gass or solid, rocky or icy. All of them spin around the sun and all have objects spinning around them, moons.
Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.
