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
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
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:
See below in bold.
Step-by-step explanation:
1. 0.00000402
= 4.02 * 10^-6 (Counting the digits after the decimal point until we get to the 4 gives us the -6).
2. 1,900,000
= 1.9 * 10^6 ( counting the number of digits after the 1 gives us 6).
Answer:
52 quarters, 42 dimes
Step-by-step explanation:
Let's call the number of quarters q, and the number of dimes d.
q+d=94
0.25q+0.1d=17.20
Subtract q from both sides of the first equation to isolate d:
d=94-q
0.25q+0.1(94-q)=17.20
0.25q+9.4-0.1q=17.20
0.15q=7.8
q=52
d=94-q=42
Hope this helps!
Answer:
Step-by-step explanation:
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