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3 years ago

Does someone know how to do math with that equation

Physics

1 answer:

mestny [16]3 years ago

8 0

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

$\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\$

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

$\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} = 0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\$

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A team of seven psychologist is study love, but each psychologist is working from a different perspective. In your opinion, what

stepladder [879]

Answer:

- Behavioral Psychologists

- Psychodynamic Psychologist

- Biological Psychologist

- Cognitive Psychologist

- Evolutionary Psychologist

- Socio - Cultural Psychologist

- Humanist Psychologist

Explained in explanation section

Explanation:

There are different classes of psychologists but for the purpose of this question we need just 7 that would fit into the study of love. These 7 will be;

1) A behavioral Psychologist might study behaviors associated with love, such as hugging, caressing or kissing.

2) A psychodynamic psychologist might study how a people's challenges with love are caused by unmindful and unresolved conflicts.

3) A biological psychologist might study which chemicals in the brain such as dopamine make people feel loved. He might also study parts of the what parts of the brain are active when people are in love.

4) A cognitive psychologist might study how people's cognitive processes will change when they are in love.

5) An evolutionary psychologist might study how certain actions or behaviors associated with love provided our forefathers with sufficient knowledge to handle their love problems.

6) A sociocultural psychologist might study how people express love in in various cultures across the world.

7) A humanistic psychologist might study the behavior of those who feel most loved and the impact it has in their lives

8 0

3 years ago

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The line

Sladkaya [172]

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/( $\frac{1}{4}$ year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

6 0

3 years ago

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m

6 0

3 years ago

How long does it take a message to travel from earth to a spacecraft at mars at its farthest from earth (about 400 million km)?

seraphim [82]

<span>A message needs to be traveled to mars from earth. Distance between Earth and Mars (given) s = 400 million km
Speed of light = 3.00—10^5 km/ sec
We know Velocity = distance / time => time = distance / velocity
Time taken t = 400 million km / 3.00—10^5 = 400 x 10^6 / 3 x 10^5 = 1333.3 sec Time taken t = 22 min</span>

5 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