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3 years ago

7

when you jump from an elevated position you usually bend your knees upon reaching the ground. by doing this, you make the time o

f the impact about 10 times as great as for a stiff legged landing. in this way, the average force your body experiences is...

1 answer:

never [62]3 years ago

8 0

about 1/10th as great by bending your legs

Time extended decreases acceleration value.

solve using Newton's 2nd and Uniform Acceleration Laws. plug in values for descending.

Vf = 0.0 m/s

Vi = 10 m/s

t = 1 s

Acceleration is given by

a = [Vf - Vi] / t

a = [ (0.0 m/s) - (10 m/s) ] / (1 s)

a = [ -10 m/s ] / (1 s)

a = -10 m/s^2

Solve the same thing, but with time 10x, so t = 10 s

a = [ (0.0 m/s) - (10 m/s) ] / (10 s)

a = [ -10 m/s ] / (10 s)

a = -1 m/s^2

Solve for first force and second force, using a mass of 100 kg

F = m * a

F = (100 kg) * (-10 m/s^2)

F = -1,000 N

F = (100 kg) * (-1 m/s^2)

F = -100 N

So divide long-force by short-force

(-100 N) / (-1,000 N) = 0.1 (which is 1/10)

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If Ike notices that there is a new moon tonight, when should he expect there to be a new moon again?

podryga [215]

Whatever phase of the moon Ike sees, he can expect to see
the same phase of moon again, after 29.53 days later.

5 0

3 years ago

Calculate the Poynting vector at the surface of the filament, associated with the static electric field producing the current an

Vesnalui [34]

We anticipate a constant Poynting vector of magnitude since the hot resistor will be emitting heat and none of the electric or magnetic fields will change over time.

S = P/A

= I2R/ 2πrL

= 332 kW/m2

Always pointing away from the wire, this Poynting vector.

<h3>What is the Poynting vector?</h3>

Describes the size and direction of the energy flow in electromagnetic waves using a Poynting vector. It bears the name of the 1884 invention of English physicist John Henry Poynting. It stands for the electromagnetic field's directional energy flux or power flow. The Poynting vector is significant in a static electromagnetic field because it determines the direction of energy flow in an electromagnetic field. This vector represents the radiation pressure of an electromagnetic wave and points in its direction of propagation.

To learn more about Poynting vector, visit:

<u>brainly.com/question/17330899</u>

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7 0

1 year ago

Which of the following are true (choose all that apply)? Sound can travel through a vacuum. -Sound can travel through water. Lig

Lunna [17]

Answer:

option (A) - false

option (B) - true

option (C) - true

option (D) - true

option (E) - true

option (F) - true

Explanation:

The sound waves are mechanical waves that means they need a medium to travel.

The light waves are non mechanical waves it means they do not need a medium to travel.

Sound cannot travel trough vacuum.

Sound can travel through air and water.

Light can travel trough vacuum and in air and in water.

7 0

2 years ago

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

3 years ago

At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

tino4ka555 [31]

Answer:

$W=\frac{p_0V_0-p_1V_1}{\gamma-1}$

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

$pV^{\gamma}=k$

where,

$p$ = pressure

$V$ = volume

$\gamma=\frac{C_p}{C_V}$

$k$ = constant

Therefore, work done by the gas during expansion is,

$W=\int\limits^{V_1}_{V_0} {p} \, dV$

$=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV$

$=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\$

(using $pV^{\gamma}=k$ )

$=\frac{p_0V_0-p_1V_1}{\gamma-1}$

4 0

3 years ago