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3 years ago

A particle has two down quarks and one up quark. What is the charge of that particle?

Physics

2 answers:

yKpoI14uk [10]3 years ago

6 0

Answer:

The charge of the particle is 0.

Explanation:

A quark is an elementary particle of matter. Quarks combine to form Hadrons.

Quarks have fractional charges.

Up, charm and top quark has $+\frac{2}{3}$ charge.

Down, strange and bottom has $-\frac{1}{3}$ charge.

A particle which has two down quarks and one up quark will have net charge:

$2 \times (-\frac{1}{3})+1(+\frac{2}{3}) = 0$

Thus, the charge of the particle would be 0. It would be a neutral particle.

Neutron is composed of two down quarks and one up quark. It is a neutral particle.

andreev551 [17]3 years ago

3 0

The particle would be a neutron so a neutral charge. You can say the charge is 0.

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3 years ago

you are assigned to do some calculations for a movie stunt that involves a car on a straight road. the road pictured above has a

Vlada [557]

Calculate the time it takes for the car to reach the beginning of the hill given that acceleration as already found in part a is 2.89 m/s²

Answer:

5.88 s

Explanation:

Using kinematic equation, v=u+at where v and u are the final and initial velocities respectively, a is acceleration and t is time.

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3 years ago

Upper A 12-pound box sits at rest on a horizontal surface, and there is friction between the box and the surface. One side of

Maurinko [17]

Answer:

The coefficient of static friction is : 0.36397

Explanation:

When we have a box on a ramp of angle $\alpha$ , and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.

In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:

$F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)$

Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:

$N=m*g*cos(\alpha)$

and the force of static friction (f) is given as the static coefficient of friction ( $\mu$ ) times the normal N:

$f=\mu *m*g*cos(\alpha)$

When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:

$f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)$

Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:

$\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)\\\mu=\frac{sin(\alpha)}{cos(\alpha)}\\\mu = tan(\alpha)\\\mu=tan(20^o)\\\mu=0.36397$

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