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2 years ago

What are cells? what is the relationship between cells and organisms?

Physics

2 answers:

Leona [35]2 years ago

7 0

Answer:

cells are the things that u are made of

Explanation:

the relationship between cells and organisms is that cells make up organisms and without cells we would not be here

<h2><u><em>hope this helps plz mark as brainliest</em></u></h2>

Natalija [7]2 years ago

7 0

Answer:

A cell Is the smallest unit of life of an organism. group of cells forms an organism

Explanation:

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

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

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

2 years ago

Read 2 more answers

Explain how its possible for a compound to have both iconic and covalent bonds.<br><br>Thanks!

Evgesh-ka [11]

The atoms which make up the ion are covalently bonded to one another. 19) It is possible for a compound to possess both ionic and covalent bonding. a. If one of the ions is polyatomic then there will be covalent bonding within it.

3 0

2 years ago

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along

babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

$\frac{y-x}{y}=\frac{6}{15}$

15 y - 15 x = 6 y

9 y = 15 x

$y=\frac{5}{3}x$

Differentiate with respect to time.

$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}$

As given, dx/dt = 4 ft/s

$\frac{dy}{dt}=\frac{5}{3}\times 4$

$\frac{dy}{dt}=\frac{20}{3}$ ft/s

6 0

3 years ago

A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b

netineya [11]

Answer:

The acceleration is $a =51945 \ km/h^2$

Explanation:

From the question we are told that

The lift up speed is $v = 147 \ km/h$

The distance covered for the take off run is $s = 208 m = 0.208 \ km$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest

$147^2 = 0^2 + 2* a* 0.208$

=> $a =51945 \ km/h^2$

6 0

2 years ago

A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

enyata [817]

Answer:

$W = 19.8 J$

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

$F = 14 lbf$

$F = 6.35 \times 9.8 N$

$F = 62.3 N$

stretch in the spring is given as

$x = 4 in = 0.1016 m$

now we will have

$F = kx$

$62.3 = k(0.1016)$

$k = 613.125 N/m$

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

$W = \frac{1}{2}kx^2$

$W = \frac{1}{2}(613.125)(0.254)^2$

$W = 19.8 J$

7 0

3 years ago

What are cells? what is the relationship between cells and organisms?​

What are cells? what is the relationship between cells and organisms?