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3 years ago

A standard for comparison that helps to ensure that the experimental result is caused by the condition being tested is the ____.

1 answer:

4 0

A standard for comparison that helps to ensure that the experimental result is caused by the condition being tested is the control.

<u>Explanation:</u>

In an experiment, we need to use a reference or standard term with which we compare our obtained results for better accuracy and precision. So there will several conditions or boundary conditions that need to include in those experiments to prevent the experiment from having a fallback or repair.

Among these several conditions, some of the conditions are mostly given to compare with the conditions occurred during the running of the experiment. And if any case, there is any change in those conditions with respect to standard conditions during the running of the experiment, then the experiment should be stopped.

So, these conditions which are standard comparison and helps to ensure that the experimental result is caused by the condition being tested and not anything else is termed as control.

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A cannon ball launched horizontally with a speed of 20m/s and a baseball dropped off a cliff and it accelerates at a rate of 10m

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If they both start from the same height, then they both hit the ground at the
same time. It makes no difference if their horizontal speeds aren't equal.
The cannon ball still accelerates downward at the same rate as the baseball.

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3 years ago

The force measuring instrument is called

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This instrument is called a spring scale.

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2 years ago

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The pressure inside a sealed container of methane gas (CH4) is 35.0 kPa. If this 80.0 L sample

STatiana [176]

Answer:

<h3>The answer is 30.43 L</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{35000 \times 80}{92000} = \frac{2800000}{92000} = \frac{700}{23} \\ = 30.434782...$

We have the final answer as

Hope this helps you

4 0

2 years ago

Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. when released, each begins to accelerate at 2

Leya [2.2K]

1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N

2) Charge

Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]

k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

charge = 0.0000001 C = 0.0001 mili C

3 0

3 years ago

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

3 years ago