At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
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.
Since heat here is conserved that means that the heat out is equal to the heat in. We use the expression Q = mC(T2-T1). We caclulate as follows:
Q absorbed = Q released
m1 C (T-T1) = -m2 C (T-T1)
C can be cancelled since they are the same substance.
m1 (T-T1) = -m2 (T-T1)
25 (T-10) = -12 (T-30)
T = 16.49 degrees Celsius
Answer:
1.024 × 10⁸ m
Explanation:
The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.
So, ω = v₀/8RE
For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have
2v₀ = Rω
substituting ω = v₀/8RE into the equation, we have
2v₀ = v₀R/8RE
dividing both sides by v₀, we have
2v₀/v₀ = R/8RE
2 = R/8RE
So, R = 2 × 8RE
R = 16RE
substituting RE = 6.4 × 10⁶ m
R = 16RE
= 16 × 6.4 × 10⁶ m
= 102.4 × 10⁶ m
= 1.024 × 10⁸ m
