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2 years ago

How many joules are there in 200kj

Physics

2 answers:

timurjin [86]2 years ago

5 0

Answer:

$2.00 * 10^5J$

Explanation:

Make use of your conversion factors:

$200KJ * \frac{1000J}{1kJ} = 200,000J = 2.00* 10 ^5J$

Or alternatively make use of your scientific notation. $kilo = 10^3$

Therefore $200kJ = 200*10^3J=2.0*10^2*10^3J = 2.0*10^5J$

Lesechka [4]2 years ago

4 0

Answer:

200,000 joules <3

Explanation:
just accept it

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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

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Given data

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function = x^2 + y^2 ≤ 6^2

to find out

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solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

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x = a cosθ

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take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

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