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3 years ago

What two things increase the gravitational pull between two objects?

2 answers:

8 0

Two things that increase the gravitational pull between two objects are decreasing the distance between the objects and increasing the mass of the objects. the force due to gravity is F = GMm/(r^2) where G is the gravitational constant, M and m are the masses of the two objects, and r is the distance between them.

stiv31 [10]3 years ago

5 0

In order to increase the gravitational forces between two objects,
you have exactly two choices:

-- Increase the product of the two masses.

-- Decrease the distance between their centers of mass.

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Your iclicker operates at a frequency of approximately 900 mhz (900x106 hz). what is the approximate wavelength of the em wave p

Umnica [9.8K]

The clicker emits EM (electromagnetic) wave which travels at the speed of light, that is
v = 3 x 10⁸ m/s

The frequency is
f = 900mHz = 9 x 10⁸ Hz

Because velocity = frequency * wavelength, the wavelength, λ, is given by
fλ = v
λ = v/f
= (3 x 10⁸ m/s) / (9 x 10⁸ 1/s)
= 1/3 m

Answer: 1/3 m

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3 years ago

Is there a sound loud enough to kill a person ?

shepuryov [24]

In space there is. Items such as a supernova will instantly end your life.

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3 years ago

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Andrei [34K]

Yes it it if u need the last 3 I can help

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A pan of ice is placed over a flame.

AveGali [126]

A. The molecules start packed together very tightly in a solid. Then when it turns to water, the molecules can move around each other freely, but still contained. When water turns to vapor, the molecules are going crazy moving around. They are not contained at all and bounce of of each other freely.

b. The temperature rises. (ice turns to water at 33 degrees and water turns to vapor at 212 degrees)

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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

4 years ago