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3 years ago

Why do objects move at constant speed

1 answer:

7 0

Most objects move at a constant speed because of friction and acceleration. The constant speed keeps them in place, and keeps a balance.

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which of these is an example of a biotic factor that could affect the size of a population A-climate b-disease c-a natural disas

andreyandreev [35.5K]

Hey there!

Biotic factors are living organisms. This includes plants, animals, and bacteria. Climate, water, and natural disasters are abiotic factors, which means they are nonliving factors that affect the environment. Disease is caused by bacteria, which is a biotic factor.

Therefore, your answer is B) Disease.

Hope this helps!

4 0

3 years ago

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

3 years ago

Read 2 more answers

Why are there multiple versions of the scientific method?

gavmur [86]

Answer:

it's because some versions have more steps and others have less

4 0

3 years ago

On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m

nika2105 [10]

Answer:

The magnitude of the torque is 263.5 N.

Explanation:

Given that,

Applied force = 31 N

Distance from the axis = 8.5 m

She applies her force perpendicularly to a line drawn from the axis of rotation

So, The angle is 90°

We need to calculate the torque

Using formula of torque

$\tau=Fd\sin\theta$

Where, F = force

d = distance

Put the value into the formula

$\tau=31\times8.5\sin90$

$\tau= 263.5\ N$

Hence, The magnitude of the torque is 263.5 N.

4 0

3 years ago

Two 0.2304 cm x 0.2304 cm square aluminum electrodes, spaced 0.5974 mm apart are connected to a 61 V battery. What is the charge

Arisa [49]

Answer:

The charge on each plate is 0.0048 nC

Explanation:

for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:

C = (A×ε)/d

=[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)

= 7.86×10^-14 F

then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:

q = C×V

= (7.86×10^-14)×(61)

= 4.80×10^-14 C

≈ 0.0048 nC

Therefore, the charge on each plate is 0.0048 nC.

3 0

3 years ago