Answer:
4.988kW
Explanation:
According to the question, energy E extracted from the ocean breaker is directly proportional to the intensity I. It can be expressed mathematically as E ∝ I
E = kI where k is the constant of proportionality.
From the formula; k = E/I
This shows that increase in energy extracted will lead to increase in its intensity and vice versa.
If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high
E = 10kW and I = 1.20m
k = 10/1.20
k = 8.33kW/m
To know how much energy E that will be produced when they are 0.600 m high, we will use the same formula
k = E/I where;
k = 8.33kW/m
I = 0.600m
E = kI
E = 8.33 × 0.6
E = 4.998kW
The device will produce energy of 4.998kW when they are 0.600m high.
Explanation:
<em>Given</em>
<em>Work </em><em>(</em><em>W</em><em>) </em><em> </em><em>=</em><em> </em><em>9</em><em>5</em><em>0</em><em> </em><em>J</em>
<em>time </em><em>(</em><em>t</em><em>) </em><em> </em><em>=</em><em> </em><em>5</em><em> </em><em>sec</em>
<em>Power </em><em>(</em><em>P) </em><em> </em><em>=</em><em> </em><em>?</em>
<em>We </em><em>know </em>
<em>p </em><em>=</em><em> </em><em>w/</em><em>t</em>
<em> </em><em> </em><em> </em><em>=</em><em> </em><em>9</em><em>5</em><em>0</em><em>/</em><em>5</em>
<em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>9</em><em>0</em><em> </em><em>watt</em>
Given:-
- Speed of the unicycle = 20 m/s
- Time taken = 15 s
To Find: Distance travelled by the unicycle.
We know,
s = vt
where,
- s = Distance travelled,
- v = Speed &
- t = Time taken.
Therefore,
s = (20 m/s)(15 s)
→ s = (20 m)(15)
→ s = 300 m (Ans.)
I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
You have to take note of the prefix because it gives you a hint of its value with respect to the base value. The base value is ratio. When the prefix is deka, that means that 1 ration is equal to 0.1 dekaration according to the metric system. The conversion is as follows:
68 rations * 0.1 dekarations/1 ration = 6.8 dekarations