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3 years ago

7

Can someone explain to me how to do this problem ?

1 answer:

valentina_108 [34]3 years ago

7 0

Answer:

The image distance is 13.3 cm ⇒ answer B

Explanation:

* Lets study the information in the problem

- The distance between the object and the convex lens is 40.0 cm

- The lens focuses light at a distance of 10.0 cm

- The equation of the image distance is $d_{i}=\frac{d_{o}f}{d_{o}-f}$

where di is the image distance, do is the object distance and f is the

distance of the focuses lens

* Lets solve the problem

∵ do = 40.0 cm

∵ f = 10.0 cm

∵ $d_{i}=\frac{d_{0}f}{d_{o}-f}$

- Substitute the values of do and f in the equation

∴ $d_{i}=\frac{(40.0)(10.0)}{(40.0)-(10.0)}=\frac{400.0}{30.0}=\frac{40}{3}$

∴ di = 13.3333 ≅ 13.3 cm

* The image distance is 13.3 cm

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Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

1)

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

2)

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

3)

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

4)

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

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