All categories

3 years ago

1) How is chemical energy transformed into potential energy?

Physics

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

Chemical potential energy is a form of potential energy related to the structural arrangement of atoms or molecules. This arrangement may be the result of chemical bonds within a molecule or otherwise. Chemical energy of a chemical substance can be transformed to other forms of energy by a chemical reaction.

Explanation:

You might be interested in

A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is

Alika [10]

Answer: 24.97 kg

Explanation:

The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:

F = G*(M1*M2)/R^2

Where G is the gravitational constant:

G = 6.67*10^-11 m^3/(kg*s^2)

In this case, we know that

R = 0.002m

F = 0.0104 N

and that M1 = M2 = M

And we want to find the value of M, then we can replace those values in the equation to get

0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2

(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2

623.69 kg^2 = M^2

√(623.69 kg^2) = M = 24.97 kg

This means that the mass of each object is 24.97 kg

6 0

3 years ago

You hear an _____ when sound bounces off a hard surface.

Aleksandr [31]

I think b but I’m not completely sure

7 0

3 years ago

Read 2 more answers

One isotope of a metallic element has the mass number 65 and 35 neutrons in the nucleus. the cation derived from the isotope has

Lina20 [59]

The mass number of an isotope is the sum of the numbers of protons and the numbers of the neutron. Given that the mass number is 65 and the number of neutrons is 35, the number of protons is 30. The atom is then Zinc (Zn). The charge is equal to +2, as it lacks 2 more electrons.

7 0

3 years ago

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.

JulijaS [17]

Answer:

D) 61 beats per minute

Explanation:

5 0

3 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

3 years ago

Read 2 more answers