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miskamm [114]
3 years ago
12

What is ham, I'm totally serious?

Physics
2 answers:
Digiron [165]3 years ago
7 0
Ham is a dead sliced pig that you put on sandwiches
Sergeeva-Olga [200]3 years ago
3 0
Ham is meat made from a pig =]
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What are the names of the ions?
aivan3 [116]

Answer:

Well for most of them you add -ide

Explanation:

4 0
3 years ago
____________ is the total energy of a system.
faust18 [17]
The correct answer is letter c. Enthalpy (H). Entrophy means the measure of the degree of disorder. Gibbs energy  formulated by Josiah Willard Gibbs and it is the energy associated with a chemical reaction to do work.Fusion is the combining light nuclei into a heavier nucleus.
8 0
4 years ago
Violet light of wavelength 427 nm ejects electrons with a maximum kinetic energy of 0.684 eV from a certain metal. What is the w
frutty [35]

Answer:

The work function of the metal is 2.226 eV.

Explanation:

Given;

wavelength of the violet light, λ = 427 nm = 427 x 10⁻⁹ m

maximum kinetic energy, K.E = 0.684 eV

The energy of the incident light is calculated as;

E = hf = \frac{hc}{\lambda} = \frac{6.626 \ \times \ 10^{-34} \ \times\ 3\ \times \ 10^8 }{427 \ \times \ 10^{-9}} = 4.655 \ \times \ 10^{-19} \ J\\\\1 \ eV = 1.6 \ \times \ 10^{-19} \ J\\\\E =( \frac{4.655 \ \times \ 10^{-19} \ J }{1.6 \ \times \ 10^{-19} \ J} ) \ eV\\\\E = 2.91 \ eV

Apply Einstein's photoelectric equation;

E = Ф + K.E

where;

Ф is the work function of the metal

Ф  = E - K.E

Ф  = 2.91 eV - 0.684 eV

Ф  = 2.226 eV.

Therefore, the work function of the metal is 2.226 eV.

5 0
3 years ago
What is formula of perimeter of half circle​
OLga [1]

1/2 × 2 × pie × radius

= pie × radius

6 0
3 years ago
A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change
aliina [53]

Answer:

\Delta p=-1.56\ kg-m/s

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

u=\sqrt{2gh}

h = 1.8 m  

u=\sqrt{2\times 9.8\times 1.8}

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

v=\sqrt{2gh'}

h' = 1.4 m  

v=-\sqrt{2\times 9.8\times 1.4}

v = -5.23 m/s

The change in the momentum of the ball is given by :

\Delta p=m(v-u)

\Delta p=0.14(-5.23-5.93)

\Delta p=-1.56\ kg-m/s

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0
4 years ago
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