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3 years ago

15

A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down

the slide's vertical distance

1 answer:

KIM [24]3 years ago

6 0

<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>

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A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of

LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

$\Psi=4RcE\pi$

Where

$\Psi =$ The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

$I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}$

Then replacing our values we have that

$I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}$

$I = 1.3*10^7 Bq$

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

$\Psi = \frac{cE}{4\pir^2}$

$\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}$

$\Psi = 2.14*10^4 MeV/cm^2.s$

The uncollided flux density at the outer surface of the tank nearest the source is $\Psi = 2.14*10^4 MeV/cm^2.s$

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2 years ago

g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo

mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D) Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 , exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0

2 years ago

By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what

Vinil7 [7]

Explanation:

1. Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}$

$V=\dfrac{h^2}{2q_pm_p\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}$

V = 45.83 volts

2. Mass of the electron, $m_p=9.1\times 10^{-31}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}$

$V=\dfrac{h^2}{2q_em_e\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}$

$V=6.92\times 10^{34}\ V$

V = 84109.27 volt

Hence, this is the required solution.

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3 years ago

A 16 foot ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 3 feet per second, how fa

JulsSmile [24]

Answer:

11.625

Explanation:

L = length of the ladder = 16 ft

$v_{y}$ = rate at which top of ladder slides down = - 3 ft/s

$v_{x}$ = rate at which bottom of ladder slides

y = distance of the top of ladder from the ground

x = distance of bottom of ladder from wall = 4 ft

Using Pythagorean theorem

L² = x² + y²

16² = 4² + y²

y = 15.5 ft

Also using Pythagorean theorem

L² = x² + y²

Taking derivative both side relative to "t"

$0 = 2x\frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t}$

$0 = x v_{x} + y v_{y}$

$0 = 4 v_{x} + (15.5) (- 3)$

$v_{x}$ = 11.625 ft/s

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3 years ago

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BlackZzzverrR [31]

Explanation:

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2 years ago