The ______________of matter depends upon how close the individual particles are together.

What separates the terrestrial planets from the jovian or gas planets in space?

Flura [38]

In our solar system, terrestrial planets are separated from the gas giants by the asteroid belt. The asteroid belt is a region in the solar system between Mars and Jupiter where asteroids are located. Gas giants do not have a solid surface and possible a small rocky core. The gas giants are Jupiter, Saturn, Uranus and Neptune. The first four planets, Mercury, Venus, Earth and Mars.

8 0

3 years ago

A block lies on a horizontal frictionless surface and

zhenek [66]

Answer:

0.1 m

Explanation:

F = Force exerted on spring = 3 N

k = Spring constant = 60 N/m

x = Displacement of the block

As the energy of the system is conserved we have

$Fx=\dfrac{1}{2}kx^2$

$\\\Rightarrow x=\dfrac{2F}{k}$

$\\\Rightarrow x=\dfrac{2\times 3}{60}$

$\\\Rightarrow x=0.1\ m$

The position of the block is 0.1 from the initial position.

6 0

3 years ago

What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t

Gekata [30.6K]

Start by facing East. Your first displacement is the vector

d₁ = (225 m) i

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

d₂ = (350 m) j

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

d₃ = (125 m) (cos(60º) i + sin(60º) j )

d₃ ≈ (62.5 m) i + (108.25 m) j

The net displacement is

d = d₁ + d₂ + d₃

d ≈ (287.5 m) i + (458.25 m) j

and its magnitude is

|| d || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0

3 years ago

What contributions did J.J. Thomson make to atomic history?

luda_lava [24]

J.J. Thomson discovered electrons and noticed that an atom can be divided. Also, he concluded atoms are made of positive cores and negatively charged particles within it.

4 0

3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago