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bonufazy [111]
2 years ago
14

A 0.040 kg ball tied to a string moves in a circle that has a radius of 0.700 m. If the ball is accelerating at 43.2 m/s2, what

is the tangential velocity of the ball?
Chemistry
1 answer:
LekaFEV [45]2 years ago
3 0

Answer: 30.24 m/s

Explanation:

Got it right on test

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What depends on the type of acid some acid you have to use plastic or aluminum to store
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3 years ago
How many grams are there in 457.25 pounds?
Ne4ueva [31]
There are 207405.111 grams in that many pounds.
7 0
3 years ago
Read 2 more answers
All solutions<br>are mixtures; but<br>all mixtures are<br>not solutions.<br>why?​
aleksklad [387]

All solutions are mixtures of two or more substances, but unless the mixture has a homogeneous distribution of solutes in the solvent, then the mixture is not a solution. Therefore, all mixtures are not solutions.

5 0
2 years ago
A standard 1.00 kg mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50
Elden [556K]

Answer:

The section of the bar is 2.92 inches.

Explanation:

Mass of the steel cut ,m = 1.00 kg = 1000 g

Volume of the steel bar = V = Area × height

Height of the of the  section of bar = h

Area of  Equilateral triangular = \frac{\sqrt{3}}{4}a^2

a = 2.50 inches

Cross sectional area of the steel mass = A

A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2

V = 2.71 inches^2\times h

Density of the steel = d =7.70 g/cm^3

1cm^3 = 0.0610237 inches^3

d=\frac{m}{v}

\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}

h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}

h = 2.92 inches

The section of the bar is 2.92 inches.

3 0
3 years ago
Please help me
Wittaler [7]

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

- log [H⁺] = pH → 6.999

A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

8 0
3 years ago
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