Answer:
v = 35.11 m / s, θ = 20º north of east
Explanation:
This is a compound problem of relative velocities, we can use the Pythagorean theorem to find the modulus of the resulting velocity
v² = v₁² + v₂²
where the eastward velocity is v₁=33 m/s and the northward velocity
v₂=12 m / s
v =
v = 35.11 m / s
the direction of this velocity can be found with the uses of trigonometry
tan θ = v₂ / v₁
θ = tan⁻¹ v₂ / v₁
θ = tan⁻¹ 12/33
θ = 20º
this angles is north of east
We have that the overhang possible for the two bricks are is mathematically given as
x=13.5m
Length of Overhang
Question Parameters:
A uniform brick of length 18 m
a maximum overhang of 9 m attained
<em>Length </em>18 m are stacked over the edge of a horizontal
surface.
Generally the equation that balances the moment is mathematically given as
![mg*(x-l/2)=mg(L-x)\\\\2x=3l/l\\\\x=3l/4\\\\x=3/4*18\\\\](https://tex.z-dn.net/?f=mg%2A%28x-l%2F2%29%3Dmg%28L-x%29%5C%5C%5C%5C2x%3D3l%2Fl%5C%5C%5C%5Cx%3D3l%2F4%5C%5C%5C%5Cx%3D3%2F4%2A18%5C%5C%5C%5C)
x=13.5m
For more information on lengths visit
brainly.com/question/8552546
Answer:
upward lift on an aircraft wing decreases as it gains altitude.
Explanation:
- The lift on an airplane wing is generated due to the the difference in the pressure on the top of the wing and the bottom of the wing in accordance with the Bernoulli's Principle.
- The pressure on the lower part of the wing is higher due to the low velocity stream of air than on the upper part of the wing.
The governing equation of the Bernoulli's Principle is:
![\frac{P}{\rho.g} +\frac{v^2}{2g} +z=constant](https://tex.z-dn.net/?f=%5Cfrac%7BP%7D%7B%5Crho.g%7D%20%2B%5Cfrac%7Bv%5E2%7D%7B2g%7D%20%2Bz%3Dconstant)
where:
P = pressure of the fluid
g = acceleration due to gravity
density of fluid
v = velocity of the fluid
z = height of fluid from the datum
<u>But the lift force on the wings depends upon several aerodynamic factors given mathematically as:</u>
![L=cl. \rho.A.\frac{v^2}{2}](https://tex.z-dn.net/?f=L%3Dcl.%20%5Crho.A.%5Cfrac%7Bv%5E2%7D%7B2%7D)
where:
cl = experimental constant
density of air
A = area of wing
v = velocity of the air
As we move up in the atmosphere the density of air reduces and thus the force of lift will eventually decrease, that is the reason why airplanes have a flight ceiling, an altitude above which it cannot fly.
Answer:
15
Explanation:
displacement = initial position - final position
Answer:
P=1.53 i +1.92 j kg.m/s
P=2.45 kg.m/s
α = 51.34
Explanation:
Given that
m=123 g = 0.123 Kg
U= 25 m/s
θ=30°
t= 0.6 s
This is the case of projectile motion
So the horizontal component of velocity U = u cosθ
u = 25 cosθ
u = 25 cos 30°
u=21.65 m/s
The vertical component of velocity U = U sinθ
Vo= U sinθ
Vo= 25 sin 30°
Vo = 12.5 m/s
We know that horizontal component of velocity of ball will remain same.So the horizontal component of momentum
Px= m u
Px= 0.143 x 12.5 kg.m/s
Px=1.53 kg.m/s
The vertical component of ball after 0.6 s
V= Vo- g t
V= 21.65 - 10 x 0.6 m/s
V= 15.65 m/s
Py= m V
Py= 0.123 x 15.65 kg.m/s
Py=1.92 kg.m/s
P=1.53 i +1.92 j kg.m/s
Magnitude P
![P=\sqrt{1.53^2+1.92^2}\ kg.m/s](https://tex.z-dn.net/?f=P%3D%5Csqrt%7B1.53%5E2%2B1.92%5E2%7D%5C%20kg.m%2Fs)
P=2.45 kg.m/s
Direction
![tan\alpha =\dfrac{P_y}{P_x}](https://tex.z-dn.net/?f=tan%5Calpha%20%3D%5Cdfrac%7BP_y%7D%7BP_x%7D)
α = 51.34° (measured from x direction)