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mixas84 [53]
3 years ago
5

If you add a small amount of hydrochloric acid to 4 liters of water, you would create a _____ solution.

Physics
1 answer:
Harman [31]3 years ago
5 0

Answer:

The answer is D.

Explanation:

An example of a weak base is ammonia. It does not contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. The position of equilibrium varies from base to base when a weak base reacts with water. The further to the left it is, the weaker the base.

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A rocket fires two engines simultaneously. One produces a thrust of 725Ndirectly forward while the other gives a 513N thrust at
faust18 [17]

The magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

<h3>What is the resultant force the two engines exert on the rocket?</h3>

The resultant force on the rocket is calculated thus:

The 513N thrust is resolved into vertical and horizontal components;

Horizontal component: 513N cos(32.4°) = 433.14 N

Vertical component: 513N sin(32.4°) = 274.88 N

Total forward force on the rocket = 725 N + 433.14 N = 1,158.14 N

Total force at right angles:

0 + 274.88 N = 274.88 N

The resultant force (F) is then given as follows:

F² = a² + b²

F² = (1158.14 N)² + (274.88 N)²

F = √1,416,847.27

F = 1,190.3

To find the direction:

tan X 274.88 N / 1,158.14 N

X = tan⁻¹ 0.237346089419241

X = 13.35°

Therefore, the magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

In conclusion, the resultant force is obtained by resolving the forces into vertical and horizontal components.

Learn more about resultant force at: brainly.com/question/17434363

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8 0
2 years ago
What is a force field
amid [387]
Chiefly in science fiction) an invisible barrier of exerted strength or impetus.
3 0
3 years ago
A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact
ArbitrLikvidat [17]

Answer:

  • 0.09 % of the original radioactive nucllde its left after 10 half-lives
  • It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0
3 years ago
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hodyreva [135]
<span>Answer: Well, let's start by finding the pressure due to the "extra" height of the mercury. p = 1.36e4 kg/m³ · (0.105m - 0.05m) · 9.8m/s² = 7330 N/m² = 7330 Pa The pressure at B is clearly p_b = p_atmos = p_gas + 7330 Pa The pressure at A is p_a = p_gas = p_atmos - 7330 Pa c) 1 atm = 101 325 Pa Then p_gas = 101325 Pa - 7330 Pa = 93 995 Pa</span>
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3 years ago
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pork, crustaceans, blood, non-halal animal-derived additives such as gelatin or suet, alcohol and any foods containing alcohol as an ingredient

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