A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
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b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
Answer:
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Step-by-step explanation:
Answer:
ok, te ayudaré
94÷4=23.5
535÷5=107
76÷3=25.33333333333
829÷4=207.25
26÷9=2.8888888889
569÷8=71.125
97÷9=10.7777777778
527÷3=175.6666666667
25÷4=6.25
534÷9=59.3333333333
58÷6=9.6666666667
37÷5=7.4
49÷2=24.5
156÷6=26
I hope it serves you, give me a crown please
<h3>
Answer: -1.5</h3>
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Explanation:
Draw a horizontal line from W to the y axis. You should arrive somewhere between -1 and -2. While this isn't exact, it looks like we should arrive right at the middle of -1 and -2; therefore we should get to -1.5
The y coordinate of W is -1.5
Keep in mind this is based on the assumption we reach the halfway point. Unfortunately, W is not on any horizontal grid lines to be able to determine exactly where W is along the y axis.
R - 10 = 18 - r
2r = 28
r = 14
