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OverLord2011 [107]
2 years ago
9

A less common type of radioactive particle emitted is the positron. If potassium-38 forms argon-38 by positron emission, the num

ber of a positron is and the charge is
Physics
1 answer:
aleksandr82 [10.1K]2 years ago
4 0
Potassium goes to argon so that a proton in the K nucleus "vanishes" into a positron (positive electron ?) but keeps the same atomic mass at 38. ? I'd say that the number of the positron was 0, and the charge was +1 electronic charge.
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Calcular la energía cinética de un cometa cuya masa es de 5×10 elevado a 31 kg y se mueve con velocidad de 216000km/h
PolarNik [594]

The kinetic energy is 9\cdot 10^{40}J.

Explanation:

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

K is the kinetic energy of the object

m is the mass of the object

v is the speed of the object

For the comet in this problem, we have:

m=5\cdot 10^{31} kg is its mass

v=216,000 km/h is the speed

First, we convert the speed  from km/h to m/s:

v=216,000 \frac{km}{h} \cdot \frac{1000 m/km}{3600 s/h}=60,000 m/s

Therefore, the kinetic energy of the comet is

K=\frac{1}{2}(5\cdot 10^{31})(60,000)^2=9\cdot 10^{40}J

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
When a hot and cold object are placed in contact, the hot one loses energy. Does this violate energy conservation? Why or why no
blsea [12.9K]

Answer:

This does not violate the conservation of energy.

Explanation:

This does not violate the conservation of energy because the hot body gives energy in the form of heat to the colder body, this second absorbs energy. This will be the case until both bodies reach the same temperature, reaching thermal equilibrium and reducing the transfer of thermal energy. In this way the energy was only transferred from one body to another but the total energy of the system (body 1 plus body 2) will be the same as in the beginning, respecting the principle of conservation of energy or also called the first principle of thermodynamics .

The part of physics that studies these processes is in turn called heat transfer or heat transfer or thermal transfer. Heat transfer occurs whenever there is a thermal gradient or when two systems with different temperatures come into contact. The process persists until thermal equilibrium is reached, that is, until temperatures are equalized. When there is a temperature difference between two objects or regions close enough, the heat transfer cannot be stopped, it can only be slowed down.

8 0
3 years ago
A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?
Anna71 [15]

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

4 0
3 years ago
A van de Graaff generator accelerates electrons so that they have energies equivalent to that attained by falling through a pote
Marat540 [252]

Answer:

Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works

answer :

If the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

Explanation:

when a Van de Graff generator is used to accelerate an electron through a PD ( potential difference ) of any value the particle ( electron )  the electron will gain energy  ( eV )  which is is equivalent in value of the PD it accelerated through

hence if the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

4 0
3 years ago
What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?
lisabon 2012 [21]
<h3>Answer:</h3>
  • E≈2,5 eV
<h3>Explanation:</h3>

_______________

λ=500 nm = 500·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

_______________

\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{500\cdot 10^{-9}\; m} =2,484\;  eV\approx \boldsymbol{2,5\; eV}

6 0
1 year ago
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