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2 years ago

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Who got the energy conversation in a system gizmo answers dm me on insta .al3xgonrockout

1 answer:

Mars2501 [29]2 years ago

3 0

Answer:

a

Explanation:

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What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?

svetoff [14.1K]

Base on my research, within 2 hours you have a number of atoms which remain.
N= N0*2^(-t/6.020 = N= N0*2^-0.33223= 07943 N0

So, the number of atoms that are being disintegrated is N0-N=N0*(1-0.79430)=0.2057 N0

It must be equal to 15 mCi = 15*3.7*10^7= 5.55*10^8 atoms

N0= 5.55*10*8/0.2057 = 2.698*10^9 atoms

Therefore, 2.698*10^9 atoms is the number of N0

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2 years ago

A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 1600 kPa. The working flui

anzhelika [568]

Answer:

a) 6498.84 kW

b) 0.51

c) 0.379

Explanation:

See the attached picture below for the solution

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2 years ago

A seamount is an isolated land mass rising from the ocean floor.<br> a. True<br> b. False

iris [78.8K]

The answer is A, True.

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2 years ago

Two forces F1 = -7.90i + 5.40j and F2 = 9.00i + 7.60j are acting on an object with a mass of m = 6.30 kg. The forces are measure

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1 year ago

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

2 years ago