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3 years ago

How do scientists learn about the brain.

2 answers:

7 0

Mainly looking at scans ansd the levels of hormons and chemicals when we use or brain for example when we use the frontal lobe diffren thsades will appear. our stress levels as well. how we react to things problems situations or questions.

Allushta [10]3 years ago

3 0

By doing scans and dissecting animals (or people who donated their body to science when they die).

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How does muscle fatigue affect the amount of work that muscles can do?

Ganezh [65]

Answer:

by straining that muscle it can slow down the amount of muscle your supposed to get

Explanation:

6 0

3 years ago

Read 2 more answers

Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c

amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

$F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F$

Then, it means that the net force acting on the test charge has a magnitude of √2F.

7 0

3 years ago

Which situation is NOT the result of an unbalanced force acting on an object? A. an object speeds up B. an object maintains spee

7nadin3 [17]

Im pretty sure its b........

5 0

3 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

4 years ago

From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 121.6 nanome

lina2011 [118]

Answer:

b) The star is moving away from us.

Explanation:

If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.

7 0

3 years ago