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astraxan [27]
3 years ago
14

A Rocket is launched and reaches a height of 72m before falling back to Earth a) What was it's take off velocity? b) What was it

's final velocity when it landed? c) What was it's velocity at the highest point?​
Physics
1 answer:
DerKrebs [107]3 years ago
4 0

Answer:

Answers are in the explanation

Explanation:

a)

vf² = vi² + 2gh

0 = vi² + 2(-9.8)(72)

vi = 37.57 m/s

b)

vf = vi + gt

0 = 37.57 - (9.8)t

t = 3.83 s

3.83 x 2 = 7.67 s

vf = vi + gt

vf = 37.57 - (9.8)(7.67)

vf = -37.60 m/s

c)

Its vertical velocity at the highest point is 0 m/s

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Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag
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Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β  ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β  =  0,5/0,7

tan β  = 0,7142    then   β = arctan 0,7142      ⇒   β = 35 ⁰

cos  β = 0,81

d = √ (0,5)²  +  (0,7)²       d1stance between charges

d = √0,25 + 0,49

d =  √0,74  m

d = 0,86  m

Now Foce between two charges  is:

F  = K* q₁*q₂/ d²      (1)

Where  K  = 9*10⁹ N*m²/C²

q₁  =  2,5* 10⁻⁹C

q₂  = 3,0*10⁻⁹C

d²  = 0,74 m²

Plugging these values in (1)

F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹   [N]

And  Fx  =  F*cos  β

Fx  =  91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹  [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹  [N]

F = 147,78*10⁻⁹ [N]

8 0
3 years ago
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