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3 years ago

14

If the focal length of a concave mirror is 18cm, find its radius of curvature.

1 answer:

Galina-37 [17]3 years ago

8 0

Given :

The focal length of a concave mirror is 18 cm.

To Find :

The radius of curvature of the concave mirror.

Solution :

We know,

$\text{Focal length}=\dfrac{\text{Radius of curvature}}{2}\\\\F=\dfrac{R}{2}\\\\R = 18\times 2\ cm\\\\R = 36 \ cm$

Therefore, the radius of curvature of concave mirror is 36 cm.

Hence, this is the required solution.

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Light incident on a Surface at an angle of 45° undergoes diffused reflection. At what angle will it reflect?

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A parcel of air is being forced up the windward side of a 7,000-foot-high mountain. The parcel has a temperature of 91.0°F at 0

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Answer:

Temperature of parcel at 1000 ft is 85.5°F

Explanation:

Temperature of parcel at 0 ft =T₁ = 91.0°F

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As (1000<5000), so using the formula for parcel below or at the LCL i.e.

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3 years ago

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Answer:

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3 years ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

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3 years ago