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nadezda [96]
3 years ago
8

Stion 3

Physics
1 answer:
GaryK [48]3 years ago
6 0

Answer:

algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium. According to the principle of moments, in equilibrium: Sum of anticlockwise Moments = Sum of clockwise moments.

Condition:

<em>For an object to be in equilibrium, it must be experiencing no acceleration.</em>

<em>Hope </em><em>it </em><em>may </em><em>helpful</em><em> to</em><em> you</em>

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The established value for the speed of light in a vacuum is 299,792,458 m/s. What is the order-of-magnitude of this number?
iogann1982 [59]
The order of magnitude is 10⁸ .
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3 years ago
Estimate the Joule-Thomson coefficient of refrigerant-134a at 0.70 MPa and 50°C. Assume the second state will be selected for a
leva [86]

Answer:

\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

Explanation:

The Joule-Thomson coefficient is the ratio of the change of temperature to the change of pressure under isoenthalpic conditions:

\mu = \left(\frac{\Delta T}{\Delta P}\right)_{h}

Initial and final properties are:

T_{1} = 50^{\textdegree}C, P_{1}=700\,kPa, h_{1}=288.54\,\frac{kJ}{kg}. Superheated Vapor.

T_{2} = 48.186^{\textdegree}C, P_{2}=600\,kPa, h_{2}=288.54\,\frac{kJ}{kg}. Superheated Vapor.

The Joule-Thomson coefficient is approximately:

\mu = \frac{50^{\textdegree}C-48.186^{\textdegree}C}{700\,kPa-600\,kPa}

\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

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4 years ago
Lightwaves travel from the air into a lens made of glass. Their velocity decreases as they enter the glass. How does this affect
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Read 2 more answers
Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially atx1 = 14.7 mm.In this confi
Sladkaya [172]

Answer:

The force magnitude is 1.75 μN acting outward from the origin towards x₂

Explanation:

The given parameters, are;

The location of q₁ = The origin

The location of q₂ = 14.7 mm from q₁

The repulsive force exerted on q₂ by q₁ = 2.62 μN

The location the particle q₂ is located to =  18.0 mm from q₁

By Coulomb's law, we have;

F = k\dfrac{q_1 \times q_2 }{r^2}

Where;

k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)

r = The distance between the particles

F = The force acting between the particles

When r = 14.7 mm F = 2.62 μN

∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸

q₂ × q₁ = 1.48 × 10⁻¹⁸ C²

When the distance is increased to 18.0 mm, we have;

F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N

∴ F = 1.75 μN.

Therefore;

The force magnitude is 1.75 μN outward from the origin towards  x₂.

8 0
4 years ago
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