Answer:
Explanation:
Let the intensity of unpolarised light be I₀ . After passing through the first polarising filter , the intensity is I₀ / 2 .
After second filter , the intensity will be I₀ / 2 x cos²45 = I₀ / 4
After third filter , the intensity will be I₀ / 4 x cos²45 = I₀ / 8 .
So,
1 / 8 the of initial light passes through the last filter .
Energy comes from the Sun, plants and animals that we eat, to provide us with nutrients we need for energy.
Well depending on what current the heater pulls im going to assume about 13, and 13A for the hair dryer, thats 26A on the 40A circuit.
I dont see how a lightbulb could overload the circuit.
Anyway, assuming the circuit is overloaded by some really big heater- the circuit would trip, the fuse would go and remain off. Most houses are fitted with seperate circuits for lights and sockets, so the light may remain on depending on the breaker board. - the reason for them all being able to run with the sudden overload may be due to a surge.
One solution to this is not to put such a large heater on the circuit with other appliances.
Another may be to dry your hair in the dark
Answer:
16.1 m/s
Explanation:
We can solve the problem by using the law of conservation of energy.
At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by
where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to
where m = 0.15 kg is the mass of the block and v is its speed.
Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:
Try looking at it like this and I'll bet you'll get it:
You're running through the boxcar on the train at 112 meters per second, but you notice that you're moving along the tracks at 210 meters per second. What must be the speed of the train along the tracks ?
