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fomenos
3 years ago
13

How to find volume of a cuboid in cm3

Physics
2 answers:
Alex73 [517]3 years ago
7 0

Answer:

<em>we</em><em> </em><em>have</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>formula</em><em> </em><em>of</em><em> </em><em>volume</em><em> </em><em>to</em><em> </em><em>find</em><em> </em><em>volume</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>cuboid</em><em>.</em><em> </em><em>(</em><em> </em><em>i.e</em><em> </em><em>v</em><em> </em><em>=</em><em> </em><em>l</em><em> </em><em>×</em><em>b</em><em> </em><em>×</em><em>h</em><em>)</em>

Explanation:

here, let your length of cuboid be x cm, breadth be y cm and height be z cm .

now, formula to find volume of cuboid = length ×

breadth × height.

so, v( volume)= l (length)× b (breadth)× h (height)

or, v= x cm × y cm × z cm

therefore, volume is xyz cm^3..... answer.

<em><u>hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

Kisachek [45]3 years ago
7 0

Answer: length times width times height

Explanation:

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3 years ago
A diffraction grating has 500 slits/mm. What is the longest wavelength of light for which there will be a third-order maximum?
Alexxandr [17]

Answer:

The longest wavelength of light  is 666.7 nm

Explanation:

The general form of the grating equation is

mλ = d(sinθi + sinθr)

where;

m is third-order maximum = 3

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d is the slit spacing (m/slit)

θi  is the incident angle

θr is the diffracted angle

Note: at longest wavelength, sinθi + sinθr = 1

λ = d/m

d = 1/500 slits/mm

λ = 1 mm/(500 *3) = 1mm/1500 = 666.7 X 10⁻⁶ mm = 666.7 nm

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8 0
4 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

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3 years ago
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Answer:

η = 2.57%

Explanation:

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