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kipiarov [429]
3 years ago
9

How many cubic objects of volume 2cm cube can be started in a room of dimension 2m by 3m by 4m​

Physics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

12,000,000 boxes

Explanation:

the volume of the room can be found by using the equation for volume of a rectangular box:V=LxWxH

where:

L=2m

W=3m

H=4m

(it doesn't really matter which is which since it is multiplication)

when we multiply our values (2m*3m*4m) we get 24cubic meters

now we need to convert cubic meters to cubic centimeters

each cubic meter is 1,000,000 cubic centimeter we multiply 24 by 1,000,000 and we get: 24,000,000 cubic centimeters (cc)

dividing 24,000,000 by 2 (since each box is 2cc) we get 12,000,000

so, we know we can fit 12,000,000, 2 cubic centimeter boxes in this room

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A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.
victus00 [196]

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5 X N₀(A)  = N₀(B)

N = N₀ e^{-\lambda t}

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ e^{-\lambda_1 t}

For B

N(B) = N(B)₀ e^{-\lambda_2 t}

N(A) = N(B) , for t = 2 h

N(A)₀ e^{-\lambda_1 t} = N(B)₀ e^{-\lambda_2 t}

N(A)₀ e^{-\lambda_1 t} = 5 x N₀(A)  e^{-\lambda_2 t}

e^{-\lambda_1 t} = 5  e^{-\lambda_2 t}

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

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For A

.77 =  .693 / λ₁

λ₁ = .9 h⁻¹

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

Putting t = 2 h , λ₁ = .9 h⁻¹

e^{\lambda_2\times  2} = 5  e^{.9\times  2}

e^{\lambda_2\times  2} = 30.25

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λ₂ = 1.7047

Half life of B = .693 / 1.7047

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3 years ago
What is the potential energy of a 2kg object placed 6m above<br> the surface of the Earth?
zalisa [80]

Answer:117.6joules

Explanation:

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PE=2 x 9.8 x 6

PE=117.6joules

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3 years ago
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