Answer:
There are six main components, or parts, of weather. They are <u>temperature, atmospheric pressure, wind, humidity, precipitation, and cloudiness</u>. Together, these components describe the weather at any given time. These changing components, along with the knowledge of atmospheric processes, help meteorologists—scientists who study weather—forecast what the weather will be in the near future.
a) 10 m/s
b) 25 m
Explanation:
a)
The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time
For the body in this problem:
u = 0 (the body starts from rest)
is the acceleration
t = 5 s is the time
So, the final velocity is

b)
In this second part, we want to calculate the distance travelled by the body.
We can do it by using another suvat equation:

where
u is the initial velocity
v is the final velocity
a is the acceleration
s is the distance travelled
Here we have
u = 0 (the body starts from rest)
is the acceleration
v = 10 m/s is the final velocity
Solving for s,

Answer:
a) 15.77 m/sec2
b) 13.3 deg
Explanation:
we are given;
Flea force = F1=1.07×10⁻5 N j
Breeze force = F2 = 1.14× 10⁻6 N (-j
mass of flea =6.0 ×10⁻7 kg
So net force on the flea=F1+F2+weight of flea=1.07×10⁻5 j +1.14× 10⁻6 i + 6.0 ×10⁻7 (-j) ×9.8= ma
==> ma = 1.07×10⁻5 j - 0.588×10⁻5 j + 0.114×10⁻5 i
==> ma= 0.482 ×10⁻5 j +0.114×10⁻5 i
==> ma = 0.114×10⁻5 i +0.482 ×10⁻5 j
== a = (0.114×10⁻5 i +0.482 ×10⁻5 j) / 6.0 ×10⁻7
==> a =
==>a= (1.9 j+8.03 i ) m/sec2
mag of a
= 15.77 m/sec2
direction angle = tan⁻1(1.9/8.03)=13.3°
Elastic collision is when kinetic energy before = kinetic energy after
Ek= 1/2mv^2
total before
Ek=1/2(2)(2.2^2) = 4.84 J
total after
Ek= 1/2(2+4)(v^2) = 3v^2
Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s
We have that the maximum rank of the kangaroo is given by:
R = v0 ^ 2 sin (2θ) / g
where,
v0 = initial velocity
θ = angle of the velocity vector formed from the horizontal
g = gravity
Clearing the speed we have:
v0 ^ 2 = (R * g) / (sin (2θ))
Substituting values
v0 = root (((11) * (9.8)) / (sin (2 (21 * (pi / 180)))))
v0 = 12.69 m / s
answer
its takeoff speed is 12.69 m / s