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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Physics

1 answer:

Varvara68 [4.7K]3 years ago

6 0

The answer to your question is,

4 kilometers north

-Mabel <3

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Two convex lenses are placed 15 cm apart. The left lens has a focal length of 10 cm, and the right lens a focal length of 5 cm.

Minchanka [31]

Answer:

The image distance from right lens is 2.86 cm and image is real.

Explanation:

Given that,

Focal length of left lens = 10 cm

Focal length of right lens = 5 cm

Distance between the lenses d= 15 cm

Object distance = 50 cm

We need to calculate the image distance from left lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-50}$

$\dfrac{1}{v}=\dfrac{3}{25}$

$v=8.33\ cm$

We need to calculate the image distance from right lens

The object distance will be

$u = 15-8.33 = 6.67\ cm$

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{-6.67}$

$\dfrac{1}{v}=\dfrac{1167}{3335}$

$v=2.86\ cm$

The image is real.

Hence, The image distance from right lens is 2.86 cm and image is real.

4 0

3 years ago

Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons

gregori [183]

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

8 0

3 years ago

A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards

Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

$Y = 1.6 X - 0.032x^2$

we know that maximum height occurs is given as

$x = -\frac{b}{2a}$

$y =- \frac{1.6}{2(-0.032)} = 25$

and maximum height is

$y = 1.6\times 25 - 0.032\times 25^2$

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

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3 years ago

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Answer:

122397.7

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Answer:

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