A "screen" or even just a set of parallel bars are highly reflective to electromagnetic waves as long as the open spaces are small compared to the wavelengths.
"Grid" dishes work fine ... with less weight and less wind resistance ... for frequencies below about 3 GHz. (Wavelengths of at least 10 cm.)
(I even worked on a microwave system in South America where huge grid dishes were used on a 90-mile link.)
When considering work, we always take the force directed along the axis of motion (in this case, the horizontal axis). If 40% of the force is directed downward, then 60% of the force is being directed horizontally, so the horizontal force is 250*0.6 = 150N. Work = Force * distance = 150N * 6.2m = 930J
You can see the periodic motion as the projection over the diameter of a point moving with a circular motion.
The Amplitude will be the radius of the circumference and ω is the angular frequency (or speed) for both motions.
In the periodic motion, you will have maximum velocity at the center and it will be zero at the extremities, where the projection changes direction, while the acceleration will be maximum at the extremities and zero at the center.
The displacement will then be:
x(t) = A · cos(ωt)
And from this (using a little bit of calculus):
v(t) = A · ω · sin<span>(ωt)
a(t) = </span><span>A · <span>ω</span>² · cos(ωt)</span>
The solution is:tan(θ) = opp / adj tan(θ) = y/x xtan(θ) = y
Find x:
x = y/tan(θ)
So x = 3/tan(π/6)
Perform implicit differentiation to get the equation:
dx/dt * tan(θ) + x * sec²(θ) * dθ/dt = dy/dt
Since altitude remains the same, dy/dt = 0. Now...
dx/dt * tan(π/6) + 3/tan(π/6) * sec²(π/4) * -π/4 = 0
changing the equation, will give us:
dx/dt = [3/tan(π/6) * sec²(π/6) * π/4} / tan(π/6) ≈ 12.83 km/min
Answer:
Explanation:
This problem bothers on the energy stored in a spring in relation to conservation of energy
Given data
Mass of block m =200g
To kg= 200/1000= 0.2kg
Spring constant k = 1.4kN/m
=1400N/m
Compression x= 10cm
In meter x=10/100 = 0.1m
Using energy considerations or energy conservation principles
The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring
Potential Energy stored in spring
P.E=1/2kx^2
Kinetic energy of the block
K.E =1/mv^2
Where v = velocity of the block
K.E=P.E (energy consideration)
1/2kx^2=1/mv^2
Kx^2= mv^2
Solving for v we have
v^2= (kx^2)/m
v^2= (1400*0.1^2)/0.2
v^2= (14)/0.2
v^2= 70
v= √70
v= 8.36m/s
a. Distance moved if the ramp exerts no force on the block
Is
S= v^2/2gsinθ
Assuming g= 9. 81m/s^2
S= (8.36)^2/2*9.81*sin60
S= 69.88/19.62*0.866
S= 69.88/16.99
S= 4.11m
