Answer:There's an increase in it potential energy
Explanation:there is an increase in it potential energy
Answer:
Explanation:
If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.
The distance between the 2 crests (or troughs) is the wavelength, λ.
We want the velocity of this wave. The equation that relates these 3 things is
and filling in:
so
v = 2.5(2.0) and
v = 5.0 m/s
Answer:
1. MOON has the greatest influence over the tides
2. Ocean currents can be caused by wind, density differences in water masses caused by temperature and salinity variations, gravity, and events such as earthquakes or storms.
Explanation:
b) by the end of the fifth day, group b will be more than one cm taller than group a. eliminate
Explanation:
The correct prediction based on the data shown is that by the end of the fifth day, group B plants will be more than 1cm taller than group A plants.
let us examine the table closely;
Day Group A (Water Only) Plant Height (cm) Group B (Water and Fertilizer)
1 2.0 2.0
2 2.2 2.3
3 2.3 2.8
4 2.5 3.2
5 2.6 3.8
We can see that by the end of the fifth day, the height differences between the two plants, 3.8 - 2.6 = 1.2cm, this is greater than 1cm. This claim from the experiment is very correct.
Other predictions are false.
learn more:
Experiment brainly.com/question/5096428
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Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
