Use the data provided to calculate the amount of heat generated for each cylinder height. Round your answers to the nearest tent
h.
100m: kJ
200 m: kJ
1000 m: kJ
100m: kJ
200 m: kJ
1000 m: kJ
2 answers:
Answer:
100m: 4.9 kJ
200m: 9.8 kJ
1000m: 49.1 kJ
Explanation:
In order to calculate this, you just have to calculate the PE of every cilynder, and as you know the PE is calculated with the next formula:
SO we just hve to insert the values into the formula to get our results, remember that the height will change with every sample because we are putting the cylinder at different heights:
100m:
200m:
1000m:
And you have the different heat generated by each cilynder at different heights.
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Answer:
The intensity of the light from the bulb would be
3.501 x W/
Explanation:
Given
The Power = 110 n W = 110 x W
the distance r = 50 mm = 50 /1000 = 0.05 m
The intensity can be obtained with the relationship below;
I = Power/area ......1
The area of the sphere would be used in this case since the bulb is spherical; A=4π
Putting it into equation 1, we have;
I = P/ 4π
I = 110 x / 4 x π x
I = 3.501 x W/
Therefore the intensity of the light from the bulb would be
3.501 x W/
Answer:
1.- para cubrir la superficie lateral 4.32 metros²
2.- Area de la base 2.15 metros²
3.- Volumen 1 m³
4.- Area total 6.47 metros²
Explanation:
El área lateral sera calcular el area de seis triangulos iguales cuya base es
0.80 de base x 6 lados = 4.80 metros perimetro de la base
4.80 perimetro de base x 1.80 arista lateral / 2 = 4.32m2
Area de la base:
Perímetro x Apotema / 2
Siendo la Apotema la altura de los triangulos que componen un hexagono calculada utilizando el teorema de pitágoras:
Apotema = 0.894427191
Area: 4.80 x 0.894427191 / 2 = 2.146625258
sumando el area de la base mas el area lateral se obtiene el area total
2.15 + 4.32 = 6.47 metros
Volumen de la pirámide:
Area de la Base x Altura / 3
2.15 x 1.40 / 3 = 1.00333 m3
<h2>The intensity of siren is 0.015 W m⁻²</h2>
Explanation:
The intensity of sound is inversely proportional to the square of distance between source and the observer .
Thus Intensity at a distance 25 m is I₁ ∝ I
Similarly the intensity at a distance 50 m is I₂ ∝ II
Dividing II by I , we have
=
=
=
Thus I₂ = x 0.06 = 0.015 W m⁻²