An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (
1 answer:
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<h2>Answer: 540 J</h2>
Explanation:
The Work done by a Force
refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
(1)
In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work performed is the product of the force exerted to push the box
by the distance traveled
.
Hence:
(2)
Answer:
The batteries would be connected in series while the bulbs would be connected in parallel
Explanation:
Power (W) = VI
where V = voltage, I = current and R = resistance
from V = IR , I = V/R
Power (W) now becomes = V (V/R) =
Power (W) =
from the above equation, power is directly dependent on voltage, hence the voltage has to be high for the power to be high and the power is also inversely dependent on the resistance (in this case the bulbs which act as the load)
- We have to batteries, when batteries are connected in series the total voltage becomes the summation of the two voltages hence giving a higher voltage and when they are connected in parallel their voltage remains the same. Since we want to get higher voltage we will connect the two batteries in series.
- we have two bulbs which are the resistance here, from the equation above the power is inversely dependent on the resistance so we would need its value to be minimal. When resistance is connected in series the resistance individual will be added to get the total resistance, hence the total resistance will be high but when the resistors are arranged in parallel you get the total resistance by applying the formula
which will give us a lower resistance. Hence we would connect the bulbs in parallel.
Take note that the power from this connection should not exceed the bulbs power rating so as to avoid damage of the bulbs.