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2 years ago

17) The charge of an anti-charm quark isapproximately equal to

Physics

1 answer:

Gnom [1K]2 years ago

7 0

The charge of an anti-charm quark is approximately equal to +5.33 ×10⁻²⁰ C.

<h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

The charge on electron is equal to e = -1.6 x 10⁻¹⁹C

The charge of an anti- charm quark is +23 times the magnitude of charge on electron.

So, charge on anti- charm quark is

q = +1/3 x ( 1.6 x 10⁻¹⁹C)

q = +5.33 ×10⁻²⁰ C

Thus, the charge of an anti-charm quark is approximately equal to +5.33 ×10⁻²⁰ C.

Learn more about electrons.

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

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3 years ago

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3 years ago

A system had 150 kj of work done on it and its internal energy increased by 60 kj. How much energy did the system gain or lose a

mina [271]

Answer:

The system loses 90 kJ of heat

Explanation:

We can answer the question by using the 1st law of thermodynamics, which states that:

$\Delta U=Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system (positive if absorbed, negative if released by the system)

$W$ is the work done by the system (positive if done by the system, negative if done by the surrounding on the system)

In this problem, we have:

$W=-150 kJ$ is the work done (negative, because it is done by the surrounding on the system)

$\Delta U=+60 kJ$ is the increase in internal energy

Using the equation above, we can find Q, the heat absorbed/released by the system:

$Q=\Delta U+W=+60 kJ+(-150 kJ)=-90 kJ$

And the negative sign means that the system has lost this heat.

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Answer: A.E.

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