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Tcecarenko [31]
3 years ago
15

50 Points Please Help Me for Physics

Physics
2 answers:
GalinKa [24]3 years ago
3 0

Explanation:

met I'd => 86189778400

pas code => g0cg6V

come on g met => spd-mwbz-een

Harman [31]3 years ago
3 0

#A

It's flat as no change can be seen on the graph of A part

#B

Its increasing as the line goes up in B part.

#C.

It's flat as no change can be seen on the graph of C part

#D

It's decreasing as the graph goes down in D part.

#E

It's decreasing as the graph goes down in E part

#F

It's decreasing as the graph goes down in F part.

#G

It's increasing as the graph goes up in G part.

Done

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A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

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2) s=Vo*t

Putting values we get

107=Vo*1.743

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3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
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