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Anestetic [448]

2 years ago

5

A 75.0-kg ice skater moving at 10.0 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters mov

e as a unit at 5.00 m/s. Suppose the average force a skater can experience without breaking a bone is 4 500 N. If the impact time is 0.100 s, does a bone break?

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

Explanation:

Momentum change for either skater is mΔv = 75.0(5.0) = 375 kg•m/s

As a change in momentum is equal to an impulse

375 = FΔt

F = 375/0.100 = 3750 N

As 3750 N < 4500 N no bones are broken.

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Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

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Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E

The electric field between the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

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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

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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:

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

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The interest rate that the Fed charges banks that borrow money from it is called A. the discount rate. B. the federal rate. C. t

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The answer is A. the discount rate.

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The basic building block of life is

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<span>The basic building block of life is molecules, more specifically macromolecules. There are four different macromolecules which could all be described as the building blocks of life, namely carbohydrates, proteins, nucleic acids and proteins. </span>

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A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

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Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

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2 years ago