The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
The
two precipitation peaks in Mbandaka during March to April and September to
November is due to the intertropical convergence zone.
Intertropical
convergence zone is a narrow zone located near the equator. It is where the
northern and southern air masses intersect which results to low atmospheric
pressure. Due to the intertropical convergence zone’s meeting of air masses,
often times the air pressure are lower which will results to colder air, or
even rainfall during the period of March to April, and most especially
September to November in Mbandaka.
<span>Since
Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named
as a Wetland of International importance, there is really a bigger chance that
this place experience above 60mm precipitation in a year, temperatures averaging
from 23 – 26 degrees Celsius.</span>
Answer:
4.14 cm.
Explanation:
Given,
For Coil 1
radius of coil, r₁ = 5.6 cm
Magnetic field, B₁ = 0.24 T
For Coil 2
radius of coil, r₂ = ?
Magnetic field, B₂ = 0.44 T
Using formula of maximum torque
Since both the coil experience same maximum torques
now,
Radius of the coil 2 is equal to 4.14 cm.
Answer:
7.3 newtons to the west
Explanation:
3.7kg × 11a - 3.7kg × ? = 3.7n
Answer:
the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94
Explanation:
Given that ;
the top speed of Cheetahs is almost 60 mph
In cornering abilities ; the maximum centripetal acceleration of a cheetah was measured to be = 19 m/s^2
The objective of this question is to determine the what minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration?
From the knowledge of Newton's Law;
we knew that ;
Force F = mass m × acceleration a
Also;
The net force 
= frictional force 
so we can say that;
m×a =
where;
the coefficient of static friction 
is:
= 1.94
Hence; the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94
