8a2-10ab+15b+10 Explaintion:
Answer:
14.85 m/s
Explanation:
From the question given above, the following data were obtained:
Height (h) of tower = 45 m
Horizontal distance (s) moved by the balloon = 45 m
Horizontal velocity (u) =?
Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:
Height (h) of tower = 45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
45 = ½ × 9.8 × t²
45 = 4.8 × t²
Divide both side by 4.9
t² = 45/4.9
Take the square root of both side
t = √(45/4.9)
t = 3.03 s
Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:
Horizontal distance (s) moved by the balloon = 45 m
Time (t) = 3.03 s
Horizontal velocity (u) =?
s = ut
45 = u × 3.03
Divide both side by 3.03
u = 45/3.03
u = 14.85 m/s
Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s
I can't see that cube from here.
But if the length of the side of the cube is ' K ' units,
then the surface area of the cube is 6K² units², and
the volume of the cube is K³ units³.
The ratio of the surface area to the volume is
(6K² units²) / (K³ units³) = (6) / (K units) .
So for example, if the side of the cube is 2 inches, then
the ratio of surface area to volume is "3 per inch".
That's the answer. I did the whole thing in order to earn
the points, but I don't expect you to understand much of it,
because I see from your username that you suck at math.
I'm sorry you decided that. Now that you've put up the
brick wall, it'll be even harder for any math to find its way
in there, and you'll miss out on a lot of the fun.
Force=A×M
10m/s×0.20kg
=2Newton
Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
