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igomit [66]
3 years ago
6

An object 2cm high is placed 3cm in front of a concave lens of focal length 2cm, find the magnification?​

Physics
1 answer:
Simora [160]3 years ago
8 0

Answer:

0.4

Explanation:

A concave lens is a diverging lens, so it will always have a negative focal length. Image distance is always negative for a concave lens because it forms virtual images.

From the lens formula;

1/f = 1/u+ 1/v

- 1/2 = 1/3 - 1/v

1/v = 1/3 + 1/2

v= 6/5

v= 1.2 cm

Magnification = image distance/object distance

Magnification = 1.2cm/3cm

Magnification = 0.4

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Answer:

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Explanation:

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Momentum= Mass time velocity, plug it into the formula,

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basically saying in simpler terms,

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3 years ago
Two projectile launchers are beside one another on level ground. Both launchers are directed at the same angle with respect to g
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(a) Projectile B will travel 4 times as far as projectile A prior to landing

Explanation:

Initial velocity = v

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R_A=\frac{v^{2}\sin 2\theta}{g}

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R_B=\frac{(2v)^{2}\sin 2\theta}{g}\\\Rightarrow R_B=\frac{4v^2\sin 2\theta}{g}

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\frac{R_A}{R_B}=\frac{\frac{v^{2}\sin 2\theta}{g}}{\frac{4v^2\sin 2\theta}{g}}

Here, the angle at which the projectiles are fired at are equal.

\frac{R_A}{R_B}=\frac{1}{4}\\\Rightarrow R_B=4R_A

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