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Vedmedyk [2.9K]
3 years ago
12

What is the charge of a lambda particle in elementary charges?

Physics
2 answers:
liberstina [14]3 years ago
7 0

Answer: lambda particle have the elementary charge as +1.

Explanation:

Lambda particles is a family of subatomic hadron particles containing one up quark, one down quark, and a third quark from a higher flavour generation.

Anit [1.1K]3 years ago
6 0
Lambda particle is the product of a proton collision with a nucleus . It was found that this particle was lives for much longer time than expected: 10-10 seconds. <span>The lambda is a </span>baryon<span> which is made up of three quarks: an up, a down and a strange quark. The charge of a lambda particle in elementary charges is positive. </span>
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If the current through a resistor is 12 A and the voltage drop is 12 V, what is the power absorbed by the resistor? Answer to th
klasskru [66]

Answer:

144 W

Explanation:

The power absorbed by a resistor is the product of the current passing through it and the voltage drop across it. Hence

P=IV

P=12\text{ A}\times12\text{ V}=144 \text{ W}

Other variations of this formula are derived using Ohm's law, V=IR

P=I^2R=\dfrac{V^2}{R}

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An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over
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Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

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Let's place the reference system at the point where the mass m1 is

     y₁ = 0

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End point, at height yf = y / 2

    E_{mf} = K₁ + U₁ + K₂ + U₂

    E_{mf} = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g y_{f} + m₂ g y_{f}

Since the masses are joined by a rope, they must have the same speed

     E_{mf} = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g y_{f}

   E_{mf}= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

How energy is conserved

   Em₀ =  E_{mf}

   2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

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   3/2 v₁² = ½ g y

   V₁ = √ (gy / 3)

5 0
3 years ago
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