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2 years ago

An antibaryon composed of two antiup quarks

Physics

1 answer:

Sveta_85 [38]2 years ago

5 0

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have $+\frac{2}{3} e$ charge where as down, strange and bottom quarks have $-\frac{1}{3}e$ charge.

The antiparticle of up quark is antiup quark and has charge $-\frac{2}{3}e$ charge.

The antiparticle of down quark is antidown quark and has charge $+\frac{1}{3}e$ charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

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

Thus, antibaryon has -1e charge.

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A planet of mass M and radius R has no atmosphere. The escape velocity at its surface is ve. An object of mass m is at rest a di

zubka84 [21]

Answer:

Explanation:

Expression for escape velocity

ve = $\sqrt{\frac{2GM}{R} }$

ve² R / 2 = GM

M is mass of the planet , R is radius of the planet .

At distance r >> R , potential energy of object

= $\frac{-GMm}{r}$

Since the object is at rest at that point , kinetic energy will be zero .

Total mechanical energy = $\frac{-GMm}{r}$ + 0 = $\frac{-GMm}{r}$

Putting the value of GM = ve² R / 2

Total mechanical energy = ve² Rm / 2 r

This mechanical energy will be conserved while falling down on the earth due to law of conservation of mechanical energy . So at surface of the earth , total mechanical energy

= ve² Rm / 2 r

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2 years ago

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

Semmy [17]

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}$

$\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}$

$\dfrac{1}{u}=\dfrac{1}{6}$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-\dfrac{v}{u}$

Put the value into the formula

$m=-\dfrac{0.24}{-6}$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=\dfrac{h'}{h}$

$h=\dfrac{0.080}{0.04}$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

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2 years ago

Read 2 more answers

The leaning tower of Pisa is about 56 meters tall. A ball released from the top takes 3.4 seconds to reach the ground. The final

geniusboy [140]

S=56, u=0, v=33, a=?, t=3.4

v=u+at
33=3.4 a
a = 9.7m/s^2

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2 years ago

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A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

likoan [24]

Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

3 0

2 years ago

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900

Sedaia [141]

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

$L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}$ .

(b) Now horizontally,

$L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.$

3 0

2 years ago