Let Karen's forward speed be considered as positive.
Therefore, before the headband is tossed backward, the speed of the headband is
V = 9 m/s
The headband is tossed backward relative to Karen at a speed of 20 m/s. Therefore the speed of the headband relative to Karen is
U = -20 m/s
The absolute speed of the headband, relative to a stationary observer is
V - U
= 9 + (-20)
= - 11 m/s
Answer:
The stationary observes the headband traveling (in the opposite direction to Karen) at a speed of 11 m/s backward.
The equation for Hall voltage Vh is:
Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip.
v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m
We input the numbers into the equation and get:
Vh= 0.25*5.6*0.0012 = 0.00168 V
The maximum Hall voltage is Vh= 0.00168 V.
<span>We can think this through intuitively. A frequency of 256 Hz means that the wave has 256 cycles each second. If the wavelength is 1.33 meters, then there are 256 of them each second. Therefore, we just need to multiply the wavelength by the frequency to find the speed of sound. (Note that the units Hz = 1 / s)
v = (frequency) x (wavelength)
v = (256 Hz) x (1.33 m)
v = 340.5 m/s
The speed of sound in the vicinity of the fork is 340.5 m/s</span>
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
Answer:
98.15 lb
Explanation:
weight of plane (W) = 5,000 lb
velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s
wing area (A) = 200 ft^{2}
aspect ratio (AR) = 8.5
Oswald efficiency factor (E) = 0.93
density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}
Drag = 0.5 x ρ x 
x A x Cd
we need to get the drag coefficient (Cd) before we can solve for the drag
Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)
where
- induced drag coefficient (Cdi) =
(take note that π is shown as n and ρ is shown as 
)
where lift coefficient (Cl)= 
=
= 0.245
therefore
induced drag coefficient (Cdi) = = 
= 0.0024
- since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
- Cd = 0.0024 + 0.0024 = 0.0048
Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.
Drag = 0.5 x ρ x x A x Cd
Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb
