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kap26 [50]
2 years ago
8

hii, just wondering how to solve this? it’s simple scientific notation but my brain just isn’t functioning today so it’s be help

ful! a brief explanation is fine<3 thanks a lot!

Physics
1 answer:
arlik [135]2 years ago
8 0
Honestly just use desmos or mathyway
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A brick of mass 2.0kg is at rest. It falls to the ground through a
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I may not have the answer so i'll just give up some hints.

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

I hope my little bit (big you may say) hint help you with your question.

5 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.
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