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3 years ago

1.concave lens

Physics

1 answer:

topjm [15]3 years ago

3 0

What to do here tell please.......

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Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let $\alpha$ be the angular acceleration of the end of the rod

Taking torque about the link, we have:

$\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)$

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

$\tau = I_{rod}\ \alpha......(ii)$

From equations (i) and (ii) we have:

$mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }$

The acceleration of the end of the rod farthest from the link is given by:

$a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}$

7 0

2 years ago

Two 60 cm parallel disks are separated by 40 cm and are aligned directly on top of each other. Both disks are black surfaces wit

Crazy boy [7]

Answer:

775.48 W

Explanation:

given,

diameter of disk = 0.6 cm

length of the disk = 0.4 m

T₁ = 450 K T₂ = 450 K T₃ = 300 K

$\dfrac{d}{r_1}=\dfrac{0.4}{0.3}$ = 1.33

now,

the value of view factor (F₁₂)corresponding to 1.33

F₁₂ = 0.265

F₁₃ = 1 - 0.265 = 0.735

now,

net rate of radiation heat transfer from the disk to the environment:

$=\dot{Q_{1-3}+Q_{2-3}} = 2 \dot{Q_{1-3}}$

= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)

= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)

= 775.48 W

Net radiation heat transfer from the disks to the environment = 775.48 W

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4 years ago