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3 years ago

True or False Babies have around 100 more bones than adults.

Physics

1 answer:

wolverine [178]3 years ago

8 0

Answer:

I would say it's true

Explanation:

when u get older we have less bones

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Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

JulijaS [17]

Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=180\°$ the angle between incident phhoton and the scatered photon.

$\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))$ (2)

$\Delta \lambda=4.86(10)^{-12}m$ (3) This is the shift in wavelength

5 0

3 years ago

An inexperienced researcher runs an experiment and sets his alpha level at .40 because he can't wait to get his firstsignificant

GuDViN [60]

A potential problem is that you are willing to accept a <u>5% </u>chance of being wrong if you reject the null hypothesis.

The significance level is the probability of rejecting the null hypothesis if it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that there is a difference when there is actually no difference. Rejecting the true null hypothesis results in a Type I error.

The smaller the value of α the more difficult it is to reject the null hypothesis. Therefore, choosing a low value for α can reduce the likelihood of Type I errors. The result here is that if the null hypothesis is false, it may be more difficult to reject using a lower value for α. The alpha value or statistical significance threshold is arbitrary. Which value to use depends on your field of study.

Learn more about The potential problems here:-brainly.com/question/21836542

#SPJ4

4 0

1 year ago

Explain what differentiates the Earth’s crust and lithosphere.

Allisa [31]

Answer:

What is the difference between the crust and lithosphere? The crust (whether continental or oceanic) is the thin layer of distinctive chemical composition overlying the ultramafic upper mantle. ... The lithosphere is the rigid outer layer of the Earth required by plate tectonic theory.

Explanation:

Hope this helps!

4 0

3 years ago

Read 2 more answers

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

3 0

2 years ago

A student tests the electrical conduction of four materials.

ella [17]

Answer:

A. aluminum, iron, and silver only

Explanation:

metal elements are great conductors; aluminum, iron, and silver are all metals.

plastic isn't an element, but it still doesn't conduct electricity very well because it isn't a metal

(so I would leave out plastic, because you can only choose 3 max based on the answer choices)

3 0

1 year ago