The amount of power change if less work is done in more time"then the amount of power will decrease".
<u>Option: B</u>
<u>Explanation:</u>
The rate of performing any work or activity by transferring amount of energy per unit time is understood as power. The unit of power is watt
Here this equation showcase that power is directly proportional to the work but dependent upon time as time is inversely proportional to the power i.e as time increases power decreases and vice versa.
This can be understood from an instance, on moving a load up a flight of stairs, the similar amount of work is done, no matter how heavy but when the work is done in a shorter period of time more power is required.
Hello there!
For this:
1). Convert 10km to meters!
2). Convert the 30 minutes into seconds!
3). Use the following formula to solve for speed!
speed= distance/time
Note: The units should automatically work out to m/s. :)
My goal is to make sure you understand the problem, which is why I won't be giving you the answer. It'll be more work now, but less work in the future! :)
Hope this helped!
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If the scientist repeats the experiment over and over and gets the same results. Also if the scientist peer reviews the experiment to make sure there is no bias in his or her results.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
