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Mekhanik [1.2K]

4 years ago

14

An object is placed 12.5 cm from a lens of focal length 22.0 cm. What is the image distance?

1 answer:

Tanzania [10]4 years ago

6 0

Answer:11.5

Explanation:

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(b) A cylinder of cross-sectional area 0.65m2 and

VLD [36.1K]

Answer:

The volume of the cavity is 0.013m^3

Explanation:

To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:

Step one:

Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.

Volume of the cylinder = 2.1 / 11.053 =0.19 $m^{3}$

Step two:

From the volume of the cylinder, we can get the radius of the cylinder.

$radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m$

Step three:

From the cross-sectional area, we can obtain the radius of the cavity.

Let the radius of the cavity be = r, while the radius of the cylinder be = R

CSA of cavity =

$\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m$

Step Four:

calculate the volume of the cavity using volume = $\pi r^2 \times h$

Recall that the cavity has the same height as the original cylinder

$volume = \pi \times 0.115^2\times 0.32= 0.013m^3$

8 0

3 years ago

HELP⚠️⚠️

alexgriva [62]

Answer:

I think u are traeling at speed of light and not ur friend

Explanation:

4 0

3 years ago

Read 2 more answers

A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

5 0

4 years ago

How are uniform circular motion maps the same as linear motion maps? Check all that apply.

Flauer [41]

Answer:

A: Time is represented by dots at 1-second intervals.

D: Velocity vectors change direction as the object’s direction changes.

Explanation:

3 0

4 years ago

Read 2 more answers

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

= 1.8 * $10^{-30}$ J

4 0

3 years ago