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>
7%
is the bigger number bc 0.7 is a decimal an it like ex.... 0.50th its like half of one whole percentage
do u see wat I mean
Answer:
m∠RQS = 72°
m∠TQS = 83°
Step-by-step explanation:
m∠RQS +m ∠TQS = m∠RQT
The two angles combine to make a larger angle
So
m∠RQS = (4x - 20)
m∠TQS = (3x + 14)
(4x - 20) + (3x + 14) = 155
Group the Xs and the constants
4x + 3x - 20 + 14 = 155
Combine like terms
7x - 6 = 155
Add 6 to both sides
7x = 161
Divide by 7 on both sides
x = 23
Check:
4(23) - 20 + 3(23) + 14 = 155
92 - 20 + 69 + 14 = 155
155 = 155
But we need to find m∠RQS and m∠TQS. So plug in x = 23 to the values.
m∠RQS = 4(23) - 20 = 72°
m∠TQS = 3(23) + 14 = 83°
Checking:
72 + 83 = 155
