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3 years ago

What is the chemical formula for potassium carbonate, which is formed from

Physics

1 answer:

Dennis_Churaev [7]3 years ago

8 0

Explanation:

Production. Potassium carbonate is prepared commercially by the reaction potassium hydroxide with carbon dioxide: 2 KOH + CO2 → K2CO3 + H2O.

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A scientist claimed that fabric A is better able to resist fire than fabric B. Which option describes an experiment that will pr

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Answer:

B. Hold each type of fabric over a candle flame and time how long it takes for the fabric to start to burn.

Explanation:

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4 years ago

If the road becomes wet or crowded, you should ____. slow down and increase your following distance All choices are incorrect. m

Vaselesa [24]

Answer:

The first one.

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3 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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3 years ago

An airplane, flying horizontally at an altitude of 3 miles, passes directly over an observer. If the constant speed of the air

natima [27]

I’m not sure if this helped you so here you go

6 0

3 years ago