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3 years ago

The ethical guidelines related to animal research dictate that

Physics

1 answer:

Zigmanuir [339]3 years ago

4 0

Answer:

Animals are treated as humanely as possible.

Explanation:

Ethical guidelines relating to animals are laid down principles or laws that guide against the maltreatment of animal in any way during or for a research. As the use of some animals for various researches can not be over emphasized, because of the close proximity between their body tissues to that of humans.

The ethical guidelines enforce treating animals as humanly as possible or achievable. This ensure appropriate handling and treatment of the animals.

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In order for work to happen you MUST have?

Nataly_w [17]

Explanation:

it requires energy

hope it helps

7 0

3 years ago

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou

True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0

3 years ago

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

magnetic field intensity, $B=0.07\ T$

kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

<em>Now, form the mathematical expression of Kinetic Energy:</em>

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

6 0

3 years ago

A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0

3 years ago

Find the wavelength of the third line in the lyman series, and identify the type of em radiation.

Natalija [7]

The wavelength of the third line in the Lyman series, and identify the type of EM radiation

In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.

1 / lambda = R(h)* ( $\frac{1}{(n1)^{2} }$ - $\frac{1}{(n2)^{2} }$ )