The question involves the concept & equations associated with projectile motion.
Given:
y₁ = 1130 ft
v₁ = +46 ft/s (note positive sign indicates upwards direction)
t = 6.0 s
g = acceleration due to gravity (assumed constant for simplicity) = -32.2 ft/s²
Of the possible equations of motion, the one we'll find useful is:
y₂ = y₁ + v₁t + 1/2gt²
We can just plug and chug to define the equation of motion:
<u><em>y = (1130 ft) + (46 ft/s)t + 1/2(-32.2 ft/s²)t²</em></u>
<em>(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)</em>
Answer:
Step-by-step explanation:
let the side of outer square=2x
then x²+x²=(20√2)²
2x²=800
x²=400
x=√400=20 in
2x=2×20=40 in
area of outer square=40²=1600 in²
area of inner square=(20√2)²=800 in²
area of shaded region A=1600-800=800 in²
Answer:
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Answer:
Step-by-step explanation:
Bob ran the first part of a 12 km race at a speed of 8 kmph. He ran the second part of the
race at 10 kmph. If his total time for the entire race was 1.74 hours, how far did he run in
the first part of the race?
First part :
Total Distance = 12 km
First part Speed = 8kmph
Let distance covered on first part = x
Second part speed = 10kmhr
(8*x)
8.545 rounded to the nearest hundredth is 8.55. I think so because the number in the hundredth place is 4, the number on the right of it is 5. 5 is equal to 5 so the 4 turns into 5
