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:
x = 42
Step-by-step explanation:
The marked angles are supplementary, so their sum is 180°.
(2x +8) +(2x +4) = 180
4x +12 = 180 . . . . . . . . . simplify
x +3 = 45 . . . . . . . divide by 4 (because we can)
x = 42 . . . . . . subtract 3
_____
<em>Additional comment</em>
A "two-step" linear equation like this one is usually solved by subtracting the unwanted constant, then dividing by the coefficient of the variable. Here, we have done those steps in reverse order. This makes the numbers smaller and eliminates the coefficient of the variable. Sometimes I find it easier to solve the equation this way.
Answer:
none of it
Step-by-step explanation:
Answer:
h =2.6
Step-by-step explanation:
2.5/6
2.5+6
=
2.6
