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Monica [59]
3 years ago
6

Differenciate between fundamentals quantities and derived quantities

Physics
2 answers:
Mandarinka [93]3 years ago
5 0
<h2>Answer:</h2><h2>Fundamental quantities are quantities which are independent on other physical qualities. </h2><h2>Derived quantities. Quantities which depends on fundamental quantities </h2>

REY [17]3 years ago
3 0

Answer:

Fundamental quantities are the base quantities of a unit system, and they are defined independent of the other...

• Derived quantities are based on fundamental quantities, and they can be given in terms of fundamental quantities.

• In SI units, derived units are often given names of people such as Newton and Joule.

Explanation:

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Elements from opposite sides of the periodic table tend to form ________.
Ivenika [448]
Ionic compounds is your answer. What happens is one atom donates electron(s) to the other atom, making one positive and the other negative. The opposite atoms attract, forming an ionic bond. 

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4 0
3 years ago
A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The
Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2

Kx^2=mv^2

v=\sqrt{\dfrac{kx^2}{m}}

By putting the values

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{1300\times 0.264^2}{25}}

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

V_f^2=V_i^2-2gh

By putting the values

0^2=1.9^2-2\times 10\times h

h=0.1805 m

h=18.05 cm

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:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \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 \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• 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}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
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