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White raven [17]
3 years ago
10

Which is the average kinetic energy of particles in an object?

Physics
1 answer:
alina1380 [7]3 years ago
5 0
The amount of heat in the body in joule
You might be interested in
Click on the boxes below to indicate the number of electrons or protons in each labeled position for a zinc atom (z=30, a=65). P
g100num [7]

Answer:

Name: Zinc

Symbol: Zn

Atomic Number: 30

Atomic Mass: 65.39 amu

Melting Point: 419.58 °C (692.73 K, 787.24396 °F)

Boiling Point: 907.0 °C (1180.15 K, 1664.6 °F)

Number of Protons/Electrons: 30

Number of Neutrons: 35

Classification: Transition metal

Crystal Structure: Hexagonal

Density at 293 K: 7.133 g/cm3

Color: bluish

('lil long, sorry)

7 0
3 years ago
What net force acting on a 14 KG wagon produces an acceleration of 1.5 MS^2?
valentina_108 [34]

21N

Explanation:

Given parameters:

Mass of wagon = 14kg

Acceleration = 1.5m/s²

Unknown:

Net force on wagon = ?

Solution:

Force is a pull or push on a body that causes a body to change its state. It is expressed as:  

                 Force = mass  x acceleration

  Force on wagon = 14 x 1.5 = 21N

Learn more:

Force brainly.com/question/10470406

#learnwithBrainly

7 0
3 years ago
11.
jolli1 [7]
<h3>Reducing Surface Area.</h3>

If the surface area becomes smaller, the pressure becomes larger.

8 0
3 years ago
Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST
Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

a^2+b^2=c^2

Plug values in

88.4^2+174^2=c^2

Simplify

7814.56+30276=c^2

Add the two values

38090.56=c^2

Take the square root of both sides

195.168\approx195.168

8 0
2 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
3 years ago
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