Please help!!A camera lens 13.3 cm from a

A 6.47-mm-high firefly sits on the axis of, and 13.1 cm in front of, the thin lens A, whose focal length is 6.19 cm. Behind lens

bagirrra123 [75]

Answer:

Explanation:

For lens A

object distance u = - 13.1 cm , focal length f = 6.19 cm

From lens formula

1/v - 1/u = 1/f

1 / v + 1/13.1 = 1/6.19

1/v = 1/6.19 - 1/13.1

= .16155 - .07633

= .08522

v = 11.7 3 cm

For lens B

object distance u = - ( 55.7 - 11.73) = - 43.97 cm , focal length f = 27.9 cm

From lens formula

1/v - 1/u = 1/f

1 / v + 1/43.97 = 1/27.9

1/v = 1/27.9 - 1/43.97

= .03584 - .022742

= .013098

v = 76.35 cm

Image will be formed 76.35 cm behind lens B .

magnification of lens system

= m₁ x m₂ , m₁ is magnification by lens A and m₂ is magnification by lens B

= (11.73 / 13.1) x (76.35 / 43.97)

= .8954 x 1.73

= 1.5547

size of image = total magnification x size of object

= 1.5547 x 6.47

= 10 cm approx. The first image will be real and inverted and second image will be erect with respect to object.

6 0

3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

The balloon is 66.62 m high

Explanation:

Combined Motion

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

2 years ago

Will give brainliest.... Physics with work please

marishachu [46]

Answer:

so first wrote down

Explanation:

that will be concluded as the answer

6 0

2 years ago

On a floor directly underneath a second-floor balcony, there are several spherical drops of blood about 7 mm in diameter. Which

IrinaK [193]

The correct answer for this question is this one: "The drops dripped from a bloody knife about 2 ft above the ground."
On a floor directly underneath a second-floor balcony, there are several spherical drops of blood about 7 mm in diameter. The statement that best accounts for the drops is that the drops dripped from a bloody knife about 2 ft above the ground.

Hope this helps answer your question and have a nice day ahead.

7 0

2 years ago

What are successfulness of the Competition policy in South Africa?

Mars2501 [29]

Answer:

The product choices along with its competitive prices were provided to the consumers. Practices such as horizontal collusion and resale price maintenance was declared unlawful in 1984. Prevention of monopoly growth was the aim of the competition policy act.

Explanation:

hope this helps you

6 0

3 years ago