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4 years ago

What Could be the control in an experiment to measure the effects of gas additives on fuel

1 answer:

8 0

Well,

A control in an experiment would basically be the "normal" version of your test subjects.

In a drug testing experiment with people, the control group would be the people who don't take the drug.

In an experiment on the effects of salt on potatoes, the control group would be a potato without salt on it.

So in an experiment to measure the effects of gas additives on fuel, the control would be fuel without additives.

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

How do you find the oscillation period in seconds for different pendulum lengths?

GalinKa [24]

To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

Where L is the length of the pendulum and g is the acceleration due to gravity.

On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

3 0

1 year ago

What are the two applications of electron beam welding

Georgia [21]

First of all, there are not <u>just</u> two applications that are solely applicable to the electron beam welding process. There are MANY.

Please visit out website at the URL below and you can click the "View Application" button under each listed Industry segment to view case studies of commonly EB welded applications.

https://www.ptreb.com/electron-beam-welding-applications

And for more general information on our welding process, we have an informational section you can peruse as well:

https://www.ptreb.com/electron-beam-welding-information

Good luck with your assignment- we are glad to hear they are teaching about EBW in high school!!!

5 0

4 years ago

According to the Periodic Table of Elements, Calcium (Ca) has:

RSB [31]

Answer:

A- 20 protons and 20 electrons

Explanation:

8 0

3 years ago

Cuales son las carreras de velocidad de medio fondo y fondo?

sammy [17]

Answer:

57490

Explanation:

58=8990

6 0

3 years ago