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3 years ago

How do you get a tachyionic particle and how are they faster than light

1 answer:

4 0

<span>Tachyons are studied in an area called particle physics, and I must say this is a bit out of my league, but I'll give you some general thoughts. Tachyons are hypothetical particles resulting from what physicists call a thought experiment. Back in the 1960s, some physicists wondered what would happen if matter could travel faster than the speed of light, something that is supposed to be impossible according to the Theory of Relativity. So these particles may or may not exist because they have not been proven or disproven by real experiment as of yet. What people have done is apply existing formulas to the unique properties of tachyons (like imaginary mass!). What comes out is a particles that go faster when they lose energy with a MINIMUM velocity of the speed of light and a maximum velocity of infinity! Hope that helps Ben, theoretical physics is a weird place and is not too far off from philosophy.</span>

You might be interested in

Find the unknown mass of the block 1 needed to balance the bar. Assume that the mass of the bar is negligible. Block 1 is locate

Natasha2012 [34]

Answer:

Incomplete question: The masses of the blocks m₂ = 1.5 kg and m₃ = 2 kg

Explanation:

Given data:

L₁ = length = 0.85 m

L₂ = 0.25 m

L₃ = 0.5 m

m₂ = 1.5 kg

m₃ = 2 kg

Question: Find the unknown mass of the block 1 needed to balance the bar, m₁ = ?

The torque is zero (intermediate point of the bar)

$-m_{1} gL_{1} +m_{2} gL_{2} +m_{3} gL_{3} =0$

Is negative because mass 1 is to the left of the coordinate system (see the diagram)

$m_{1} =\frac{m_{2} gL_{2}+m_{3} gL_{3} }{L_{1} } =\frac{(1.5*0.25)+(2*0.5)}{0.85} =1.6176kg$

4 0

4 years ago

A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri

san4es73 [151]

Answer:

T = 8.55 N

Explanation:

When string makes an angle 40 degree with the vertical then it will have two forces on it

1) gravitational force (mg)

2) Tension force in string (T)

now we know that net force towards the center of the path is known as centripetal force and it is given as

$T - mg cos40 = F_c$

$T - (0.40\times 9.8)cos40 = \frac{mv^2}{L}$

$T = 3 + \frac{0.40\times 5^2}{1.8}$

$T = 3 + 5.55$

$T = 8.55 N$

6 0

3 years ago

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

$P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})$

$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

$V=3.475ft^3/s$

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

$m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s$

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

6 0

3 years ago

What is frictional less pulley

Talja [164]

Answer:

Conditions under which the belt and pulleys are operating – The friction between the belt and pulley may decrease substantially if the belt happens to be muddy or wet, as it may act as a lubricant between the surfaces.

Explanation:

I hope that this would be helpful

7 0

3 years ago

Suppose a grower sprays (8.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water

____ [38]

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

<h3>Heat released by the water when it freezes</h3>

The heat released by the water when it freezes is calculated as follows;

Q = mcΔФ

where;

m is mass of water
c is specific heat capacity of water
ΔФ is change in temperature = Фf - Фi

when water freezes, the temperature, Фf = 0 °C

Q = 82 x 4200 x (0 - 0)

Q = 0

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

Learn more about heat flow here: brainly.com/question/14437874

6 0

2 years ago