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3 years ago

A ray of light parallel to the principal axis of spherical mirror

Physics

1 answer:

lesya692 [45]3 years ago

6 0

Answer:

This question is incomplete but the completed question is below

A ray of light parallel to the principal axis of a spherical mirror passes through its _____ after reflection.

(a) focus (b) pole (c) centre of curvature (d) imaginary centre of curvature

The correct option is (a)

Explanation:

After reflection, a ray of light will usually appear to diverge after they hit a spherical mirror, extending behind the mirror to meet at a point (called the focus or focal point) between the pole and the centre of curvature of this mirror. Hence, option a is the correct option.

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Ilia_Sergeevich [38]

Use the formula,

$\Delta x=v_it+\dfrac12at^2$

where $\Delta x$ is the cart's displacement (from the origin), $v_i$ is its initial speed, $a$ is its acceleration, and $t$ is time.

$\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2$

$\implies\boxed{\Delta x=30\,\mathrm m}$

Alternatively, since acceleration is constant, we have

$\dfrac{v_f+v_i}2=\dfrac{\Delta x}t$

That is, we have these two equivalent expressions for average velocity, where $v_f$ is the cart's final velocity. Solve for $\Delta x$ :

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3 years ago

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In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

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3 years ago

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Darina [25.2K]

<span>D.) The force and displacement are in the same direction.

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dolphi86 [110]

Answer:

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Explanation:

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4 years ago