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:
Step-by-step explanation:
<u>Formula for area of rectangle:</u>
= length × width
<u>Area of the smaller rectangle:</u>
length = 7
width = 6
= 7 × 6
= 42
<u>Area of the larger rectangle:</u>
length = 7 + 4 = 11
width = 6 + 2 = 8
= 11 × 8
= 88
<u>Area of the shaded region:</u>
= Area of the larger rectangle - Area of the smaller rectangle
= 88 - 42
= 46
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
The rope is not long enough a meter is roughly 3 feet 3 inches totalling out to 98 feet and a few inches
I think answer should be c. Please give me brainlest let me know if it’s correct or not okay thanks bye
By rounding to the nearest one it means round to the nearest whole number. So basically, the number 0.7414 rounded to the nearest whole number would be 1.
