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3 years ago

What is meant by a quark’s “flavor”?

Physics

2 answers:

Viktor [21]3 years ago

7 0

Answer:

Explanation:

because a quark have many kinds of flavor

aleksandr82 [10.1K]3 years ago

7 0

I need the 5 points sorry for the inconvenience

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Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

$\lambda = 656.46 \ nm$

$\Delta \lambda = 0.04$

$c=3\times 10^8$

As we know,

⇒ $\frac{\Delta \lambda}{\lambda} =\frac{v}{c}$

On substituting the values, we get

⇒ $\frac{656.46}{0.04} =\frac{v}{3\times 10^8}$

⇒ $v=\frac{656.46}{0.04} (3\times 10^8)$

⇒ $=16411.5\times 3\times 10^8$

⇒ $=18280 \ m/s$

or,

⇒ $=18 \ km/s$

8 0

2 years ago

A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?

jekas [21]

A reference point would be something not on the ship which could be used to calculate distance traveled.

Answer: C.) A lighthouse on a nearby Island

7 0

2 years ago

Read 2 more answers

In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

galina1969 [7]

Answer:

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

5 0

2 years ago

Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro

aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

$S = \frac{Q}{T}$

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

$\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}$

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy $\Delta S_{hot}$ is smaller than $\Delta S_{cold}$