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3 years ago

When the radius of curvature of the lens increases, what happens to the focal length of the lens

Physics

1 answer:

Basile [38]3 years ago

4 0

Answer

Equation of lens law

$\dfrac{1}{f}=(n -1 )(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

where f is the focal length of the lens

n is the refractive index of the lens

R_1 and R_2 is the radius of curvature of the lens.

From the above expression we can see that focal length of the lens is directly proportional to radius of curvature.

hence, when the radius of curvature of the lens increases the focal length of the lens will also increase.

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On arrival at a vehicle collision, you observe a small fire in the engine compartment. A bystander is attempting to smother the

mina [271]

Answer:

<em>Aim at the base of the fire and use short bursts until the fire is out.</em>

Explanation:

Fire extinguishers use CO2 (Carbondioxide) as the extinguishing agent. This is because CO2 is denser than air, and does not support combustion.

Aiming at the base of the fire causes the CO2 to fall on the base of the fire, where the source of the fire is, trapping it, and preventing it from further reacting with air in a combustion reaction. Also, the short burst creates a strong wind that forces the flame to blow out.

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3 years ago

A uniform electric field of magnitude 4.6 ✕ 104 N/C is perpendicular to a square sheet with sides 3.0 m long. What is the electr

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Answer:

41.4* 10^4 N.m^2/C

Explanation:

given:

E= 4.6 * 10^4 N/C

electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field

then electric flux = ∫ E*n dA

= ∫ 4.6 * 10^4 * 3*3

= 41.4* 10^4 N.m^2/C

3 0

4 years ago

Read 2 more answers

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

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3 years ago

A ______________ is very general in nature, while a ________________ specifies what we want to study more specifically, suggesti

vova2212 [387]

Answer:

research topic and research question (hypothesis)

Explanation:

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3 years ago

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

fiasKO [112]

Explanation:

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h = 2.3 m

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So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

8 0

3 years ago