Answer:
There is no attatchment
Step-by-step explanation:
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: 473.4
Step-by-step explanation: if you can't solve it, break the first number apart, multiply both numbers, then add them together. That way there not such big numbers :)
Answer: -8.3 + 9.2 - 4.4 + 3.7
= -8.3 + 9.2 + (-4.4) + 3.7 {Additive inverse}
= -8.3 + (-4.4) + 9.2 + 3.7 {Commutative Property}
= [ -8.3 + (-4.4] + [ 9.2 + 3.7] {Associative property}
= - 12.7 + 12.9
= 0.2
Step-by-step explanation: I copied it from someone hope it helps though! :D
Answer:
wheres the question??
Step-by-step explanation:
