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3 years ago

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The superheroine Xanaxa, who has a mass of 67.5 kg , is pursuing the 70.3 kg archvillain Lexlax. She leaps from the ground to th

e top of a 179 m high building then dives off it and comes to rest at the bottom of a 15.9 m deep excavation where she finds Lexlax and neutralizes him. Does all this bring about a net gain or a net loss of gravitational potential energy?

1 answer:

Paladinen [302]3 years ago

4 0

Answer:

there is a net loss of potential energy is Ut = -0.11 10⁵ J

Explanation:

The gravitational potential energy is

ΔU = mg (y2- y1)

Where y1 is the initial position where the energy is defined as zero, in general in problems on the Earth's surface the reference system is taken at the lowest point, whereby y1 = 0 and consequently U1 = 0, this It is a matter of comfort since what matters are the changes in energy and not its value at one point explained.

Let's apply these concepts to our problem.

Jump from the ground to 179 m, let's calculate the energy change

ΔU1 = mg (y2-y1)

ΔU1 = 67.5 9.8 (179 -0)

ΔU1 = 1.18 10⁵ J gains energy

Dives up to Y3 = -15.9m below ground, therefore, this height is negative

The total height is y = 179 - (-15.9) = 194.9 m

ΔU2 = mg (y3-y2)

These are the points where the dive begins and where it arrives

ΔDU2 = 67.5 9.8 (- 194.9)

ΔU2 = -1.29 105 J loses energy

The total change in potential energy is

Ut = ΔU1 + ΔU2

Ut = 1.18 10⁵ - 1.29 10⁵

Ut = -0.11 10⁵ J

Therefore, there is a net loss of potential energy

You can also calculate ΔU = mg (Δy)

Where ΔY is the change in body position, that is -15.9 m

ΔU = 67.5 9.8 (-15.9)

ΔU = -1.1 10⁴ J = -0.11 10⁵ J

You can see that it is the same value.

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light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$

Explanation:

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$t = 32,000,000 s$ --- time

Required

The distance traveled

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So, we have:

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3 years ago

Ask Your Teacher The height h in feet reached by a dolphin t seconds after breaking the surface of the water is given by h = −16

kirza4 [7]

Answer:

1 second

Explanation:

h = −16t² + 32t

When, h = 16

16 = −16t² + 32t

Divide each of the numbers by 16

1 = -1t² + 2t

Rearrange the equation

1t²-2t+1 = 0

Solving by the quadratic formula, we get

$t=\frac{-(-2)\pm \sqrt{(-2)^2-4\times 1\times 1}}{2\times 1}\\\Rightarrow t=\frac{2\pm 0}{2}\\\Rightarrow t=1\ s$

So, time taken by the dolphin to jump out of the water and touch the trainer's hand is 1 second.

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4 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:

$E_{e}=37.937keV$

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3 years ago

Please help I’m so confused

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<em>1.wavelength</em>

<em>2.trough</em>

<em>3.amplitude</em>

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