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3 years ago

What is the difference in quark model of a meson and a baryon?

1 answer:

3 0

Mesons are made of a valence quark−antiquark pair (thus have a baryon number of 0), while baryons are made of three quarks (thus have a baryon number of 1).

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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

3 years ago

A solid sphere is placed on a frictionless floor in a very long corridor and is given a quick push so that it begins to slide, w

prohojiy [21]

I think it’s D I hope it helps

8 0

3 years ago

Little Timmy wants to measure how tall his house is. He doesn’t have a tape measure but does have a stopwatch. He recruits Big B

Komok [63]

The distance covered by an object accelerating from rest is

D = (1/2) · (acceleration) · (time)² .

In this particular case, 'acceleration' is 9.8 m/s² ... due to gravity.

D = (1/2) · (9.8 m/s²) · (1.67 s)²

D = (4.9 m/s²) · (2.789 s²)

D = 13.67 meters

6 0

3 years ago

a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi

Mars2501 [29]

The radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

r/10 = tan [arctan(1/2)] = tan (1/2)

Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = $\sqrt{5} -2$

Hence, r = $10\sqrt{5} -20$

So, the radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Learn more about radius (r) of the sphere here;

brainly.com/question/14100787

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5 0

1 year ago

A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction

IrinaVladis [17]

The coefficient of friction between the road and the car's tire is determined as 0.78.

<h3>Acceleration of the car</h3>

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

u is the initial velocity = 97 km/h = 26.94 m/s

a = (26.94)²/(2 x 47)

a = 7.72 m/s²

<h3>Coefficient of friction</h3>

μ = a/g

μ = (7.72)/9.8

μ = 0.78

Learn more about coefficient of friction here: brainly.com/question/14121363

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5 0

2 years ago

What is the difference in quark model of a meson and a baryon? ​

What is the difference in quark model of a meson and a baryon?