Answer:
1170 m
Explanation:
Given:
a = 3.30 m/s²
v₀ = 0 m/s
v = 88.0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)
x = 1173.33 m
Rounded to 3 sig-figs, the runway must be at least 1170 meters long.
Answer:
Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.
Answer:
They will meet at a distance of 7.57 m
Given:
Initial velocity of policeman in the x- direction, 
The distance between the buildings, 
The building is lower by a height, h = 2.5 m
Solution:
Now,
When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.
Thus
From the second eqn of motion, we can write:
t = 0.707 s
Now,
When the policeman was chasing across:
The distance they will meet at:
9.57 - 2.0 = 7.57 m
The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
Learn more about distance here: brainly.com/question/4931057
#SPJ1
1) <span>yes;2
6*2=12
12*2=24
24*2=48
2)</span><span>Next Term (or nth term) = ar^n-1
</span>
a = first term, i.e. 5
<span>r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 </span>
<span>n = .. </span>
<span>as you already have 1st , 2nd and 3rd terms</span>
<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
<span>= 5*27 </span>
<span>T4 = 135
</span>T5= ar^n-1
<span>= 5*3^5-1 </span>
<span>= 5*3^4 </span>
<span>= 5*81 </span>
<span>T5 = 405 </span>
