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ArbitrLikvidat [17]
3 years ago
8

At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is

150 ft/s. determine the pressure at a stagnation
Physics
1 answer:
Nastasia [14]3 years ago
5 0
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

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amm1812

Answer:

resistance of a conductor increases

Explanation:

The resistance of conductors is directly proportional to the temperature of the conductor. This implies that when the temperature of the conductor is increased, the resistance of the conductor increases likewise.

This is applied in the resistance thermometer. Resistance thermometers are useful for accurate temperature measurements at very high or very low temperatures.

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3 years ago
What does the speed of sound depend on?
melamori03 [73]
Assuming it's an ideal material, the answer is A.

The speed of sound depends on the medium that the sound travels through.
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A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

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Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0
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Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter
valentinak56 [21]

Wee can use here kinematics

as we know that

y = v*t + \frac{1}{2} at^2

for shorter tree we know that

y = 0 + \frac{1}{2}*9.8 * 2^2

y = 19.6 meter

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

y = v*t + \frac{1}{2} at^2

39.2 = 0 + \frac{1}{2}*9.8 * t^2

t = 2.83 s

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A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is
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Answer:

(a)  Length =136.58 m

(b)  T=5995 N

Explanation:

for the glider in the back

T - 1900 = 700 a

for the glider in front

12000-T -1900 = 700a

add equations

12000-3800 = 1400 a

a=5.85 m/s^2

v^2 = v0^2 + 2 a x

40^2 = 2*5.85*x

Length =136.58 m

b) plug the a back into one of the previous formula

T -  1900 = 700*5.85

T=5995 N

8 0
3 years ago
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