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Allushta [10]
3 years ago
11

3. If an undetected air bubble was trapped inside the gas collection tube, how would this affect your calculated percent yield?

Explain your answer.
Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0
When an air bubble gets trapped inside the gas collection tube, then the gas that is actually collected becomes impure. This will definitely give an inaccurate result and the calculated percentage yield will be wrong. I hope that this is the answer that you were looking for and it has come to your help.
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Find the equivalent resistance of this parallel circuit with two strands.
svlad2 [7]
In a parallel circuit, the total resistance calculated from the individual resistances is computed from the formula: 1/Rt = 1/R1 + 1/R2. substituting R1 and R2, then 
1/Rt = 1/7 + 1/49 
1/Rt = 1/6.125 = 1/ 49/8
Rt = 49/8 <span>Ω

The total resistance hence is </span>49/8 Ω
6 0
3 years ago
A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. Th
LiRa [457]

Answer:

A) total intensity = 5x10^-4W/m2

B) total intensity = 5.34x10^-4W/m2

Explanation:

Detailed explanation and calculation is shown in the image below

8 0
3 years ago
Please answer this for 15 points
lisov135 [29]

Answer:

?

Explanation:

4 0
3 years ago
An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

4 0
3 years ago
An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons
Nady [450]

To solve this problem it is necessary to apply the concepts related to the conservation of energy, specifically the potential elastic energy against the kinetic energy of the body.

By definition this could be described as

PE = KE

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

Where

k = Spring constant

x = Displacement

m = mass

v = Velocity

This point is basically telling us that all the energy in charge of compressing the spring is transformed into the energy that allows the 'impulse' seen in terms of body speed.

If we rearrange the equation to find v we have

v = \sqrt{\frac{kx^2}{m}}

Our values are given as

m = 1000kg

k = 5.75*10^6N/m

x = 3.12*10^{-2}m

Replacing at our equation we have then,

v = \sqrt{\frac{kx^2}{m}}

v = \sqrt{\frac{(5.75*10^6)(3.12*10^{-2})^2}{1000}}

v = 2.3658m/s

Therefore he speed of the car before impact, assuming no energy is lost in the collision with the wall is 2.37m/s

4 0
3 years ago
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