Answer:
I_FWHW = 3.2 μW / m²
Explanation:
In the analysis of optics and electricity a very useful magnitude is the width at half height (FWHW) and the intensity at this height, which is given by
I_FWHW = I₀ / 2
corresponds to the width of the line for this intensity.
In this case they give the maximum intensity for which
I_FWHW = 6.2 / 2
I_FWHW = 3.2 μW / m²
You do not give more data in your exercise, but the most interesting calculation is to find the angle values for which you have this intensity since it is this range is 50% of the energy of the system, have I write the equation for this calculation
I = Io cos² x₁ (sin x / x)²
x₁ = π d sin θ /λ
x = π b sin θ /λ
where d is the separation of the slits and b the width of each slit
(A) power = 0.208 kW = 208 watts
(B) energy = 6.6 x 10^{9} joules
Explanation:
energy consumed per day = 5 kWh
(a) find the power consumed in a day
1 day = 24 hours
power = \frac{energy}{time}
power = \frac{5}{24}
power = 0.208 kW = 208 watts
(b) find the energy consumed in a year
assuming it is not a leap year and number of days = 365 days
1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds
energy = power x time
energy = 208 x 31,536,000
energy = 6.6 x 10^{9} joules
Answer:
An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes.
Explanation:
<span>The current from the power source will not change when the second light bulb is attached. This is because the current is kept on a single path, it will return to the power source at the same value it left the source at.</span>
Answer:
80%
Explanation:
The thermal energy output by the oven is 800 J which is the useful energy that can be utilized for cooking.
The electrical energy input to the oven is 1000 J.
The efficiency of the oven =
(Useful energy output)/(Total energy input to the oven) x 100
Hence, the efficiency of the oven is 80%.
