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stepan [7]
3 years ago
6

An astronaut having mass 320 kg with equipment included is attempting an untethered space walk. The astronaut is initially at re

st with respect to the nearby spacecraft, but discharges 30.0 g of nitrogen gas at 900.0 m/s from a portable propulsion unit. How fast is the astronaut moving after doing that?
84 m/s

27 m/s

0.36 m/s

0.084 m/s
Physics
1 answer:
ExtremeBDS [4]3 years ago
7 0
This can be solved using momentum balance, since momentum is conserved, the momentum at point 1 is equal to the momentum of point 2. momentum = mass x velocity
m1v1 = m2v2
(0.03kg x 900 m/s ) = 320(v2)
v2 = 27 / 320
v2 = 0.084 m/s is the speed of the astronaut
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Answer:

Proof in explanataion

Explanation:

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i)

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ii)

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F = Force = Newton = kg.m/s² = [MLT⁻²]

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[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\

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The initial concentration of acid ha in solution is 0.39 m. if the ph of the solution at equilibrium is 0.76, what is the percen
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The percent ionization of the acid is 44.56%

<h3>How can we calculate the percent ionization of the acid?</h3>

To calculate the percent ionization of the acid we are using the formula,

The H⁺ ion concentration [H⁺] = C x,

where, we are given,

C= concentration of the acid.

=0.39 M

x= degree of dissociation of the acid.

And one more thing we are given that, the pH of the acid=0.76.

So from the above statement we can say that,

pH = - log [H⁺]

Or,0.76 = -log [H⁺]

Or, log [H⁺] = -0.76

Or, [H⁺] = antilog -0.76

Or,[H⁺]= 10^-0.76

Or,[H⁺]=0.1738.

Now from the above calculation we know, the H⁺ ion concentration= 0.1738 M.

Now we put the known values in the above equation,

[H+]= Cx

Or,0.1739= 0.39 x

Or, x= 0.4459

From the above calculation we can conclude that the percent Ionization of the acid= 0.4459 X 100= 44.59%≈45%

Learn more about Ionization:

brainly.com/question/1445179

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