How does a wind turbine make electricity?

A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

goblinko [34]

Oh my lord lol I was do ready to help then I saw numbers

4 0

3 years ago

Euglena are _______.<br><br> A heterotrophs<br><br> B autotrophs

zysi [14]

The Euglena is unique in that it is both heterotrophic (must consume food) and autotrophic (can make its own food).

6 0

3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

3 years ago

A 2.26 kg book is dropped from a height of 1.5 m. What is its acceleration?

cluponka [151]

It should be 3.39 when u multiply 2.26 from 1.5 it gone give u 3.39 it be better if u learn the triangle

3 0

3 years ago

David lifts a book 1.2 meters from the floor to the top of his desk. If the book weighed 0.50 kg, what is the gravitational pote

Anika [276]

Answer:

5.886 J

Explanation:

Given:

The mass of the book is, $m=0.5$ kg

Height of lift is, $\Delta h=1.2$ m

Acceleration due to gravity is, $g=9.81$ m/s²

Now, gain in gravitational potential energy is a function of change in position and is given as:

$\Delta PE=mg\Delta h$

Here, $\Delta PE$ is the change in gravitational potential energy.

Plug in 0.5 kg for $m$ , 9.81 for $g$ and 1.2 for $\Delta h$ . Solve for $\Delta PE$

$\Delta PE=0.50\times 9.81\times 1.2=5.886\ J$

Therefore, the gain in gravitational energy of the book is 5.886 J.

7 0

3 years ago