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Mazyrski [523]
3 years ago
6

An important step in science is supporting a theory or idea without data. The questions we ask determine the type of data we col

lect.
In the warm up, you reviewed the question to calculate kenetic energy. What question could you ask about kenetic energy which will include the variable that will affect it?
Chemistry
1 answer:
san4es73 [151]3 years ago
3 0

Answer:

What will happen to Uk if you double the mass?

Explanation:

Uk = 0.5 * m * v²

You see that both m and v are variable, which means that both m and v can be any number. Regardless of the numbers you put in for m or v, the formula to calculate the kinetic energy (Uk) remains valid.

You could ask

1. What will happen to Uk if you double the mass?

2. What will happen to Uk if you double the velocity?

please see and understand(!) that the relationship between Uk an v² is indeed the velocity squared....

EXTRA

Uk = 0.5 * m * (v)²

Suppose the m = 3kg and velocity = 5 m/s

What is the Uk?

Well if you know the formula you can use your calculator to find out:

Uk = 0.5 * m * (v)²

Uk = 0.5 * 3 * (5)²

Uk = 0.5 * 3 * 25

Uk = 37.5 kgm/s²

Again you ask what will happen to Uk if you double the velocity?

At first it was 5 m/s and now it doubles, which means it now has that value *2

The new velocity is 5 *2 = 10 m/s

Uk = 0.5 * m * (v)²

Uk = 0.5 * 3 * (10)²

Uk = 0.5 * 3 * 100

Uk = 150 kgm/s²

150 = 4 * 37.5

So now you see that if you double your velocity, the Uk will be 2² = 4 times as big !

You might be interested in
How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina
liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.</em>

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 =  [A⁻] / [HA] <em>(1)</em>

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 <em>(2)</em>

Replacing (1) in (2):

0.3373 =  0.08255 - [HA] / [HA]

0.3373[HA] =  0.08255 - [HA]

1.3373[HA] = 0.08255

<em>[HA] = 0.06173 moles</em>

Thus:

[A⁻]  = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, <em>you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. </em>As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

3 0
3 years ago
A certain automobile contains 4 tires, 2 headlights, 1 steering wheel,
Step2247 [10]

Answer:

143 parts I think probably not

5 0
3 years ago
Jai besoi daide
JulijaS [17]

Answer:

i dont know what you are saying

Explanation:

????????????

4 0
2 years ago
What changes are observed on heat capacity ratio and output temperatures by increasing the specific heat capacity of both the fl
BlackZzzverrR [31]

Answer:

b- The heat capacity ratio increases but output temperature don’t change

Explanation:

The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.

Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.

On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.

8 0
3 years ago
​29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.
arsen [322]

The answer for the following mention bellow.

  • <u><em>Therefore the final temperature of the gas is 260 k</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 150.0 kPa

Final pressure (P_{2}) = 210.0 kPa

Initial volume (V_{1}) = 1.75 L

Final volume (V_{2}) = 1.30 L

Initial temperature (T_{1}) = -23°C = 250 k

To find:

Final temperature (T_{2})

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas  constant

T represents the temperature of the gas

We know;

\frac{P*V}{T} = constant

\frac{P_{1} }{P_{2} } × \frac{V_{1} }{V_{2} } = \frac{T_{1} }{T_{2} }

Where;

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(T_{1}) represents the initial temperature of the gas

(T_{2}) represents the final temperature of the gas

So;

\frac{150 * 1.75}{210 * 1.30} = \frac{260}{T_{2} }

(T_{2}) =260 k

<u><em>Therefore the final temperature of the gas is 260 k</em></u>

<u><em></em></u>

3 0
3 years ago
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