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2 years ago

I need help I’ll give you 15 points for both of them

Physics

1 answer:

Simora [160]2 years ago

5 0

Answer:

Point c

Explanation:

Because of gravity

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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

3 years ago

How does an electromagnet become permanent

love history [14]

<span>A moving electrical charge produces a magnetic field and a moving magnetic field produces an electrical field. An electromagnet works by coiling a bunch of wire and spinning a couple of magnets around that wire at high speeds. When this occurs the magnets induce an electric current in the wire and hence the electricity production. Once the magnets stop spinning, the induced electrical field dissipates and the current stops flowing through the wire.

</span>

6 0

2 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

2 years ago

Read 2 more answers

Which of the following is true of high clouds?

Wewaii [24]

I’m pretty sure it’s B

3 0

2 years ago

1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer

Aloiza [94]

Answer:

Following are the solution to the given question:

Explanation:

Its strength from both charges is equivalent or identical. The power is equal. And it is passed down

$F=\frac{kq_1q_2}{r^2}$

Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.

$|F_{12}| = |F_{21}|$

7 0

2 years ago