Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under 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):
The velocity at the stagnation point is zero.
The density stays constant.
Let p₂ denote the pressure at the stagnation location.
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
Thus, the answer is 2.2 psi
Answer:
The duration, t = 3.53 seconds
Explanation:
The following information is provided:
The equation to calculate the time t is expressed as:
...... (1)
Where
s denotes the distance in feet
We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet
Substituting the value of s into equation (1) yields:

t = 3.53 seconds
Thus, the time taken by the object is 3.53 seconds, which provides the required answer.
Response:
The resistivity can be expressed as 
Clarification:
According to the information provided,
The value of the electric field measures 
The density of current is given as 
Typically, resistivity is represented in mathematical terms as

by inserting values


The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.
Answer:
35.79 meters
Explanation:
We have an archer, and there is a target. Denote the distance between them as d.
The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time
. Thus, we can derive:
.
Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:
.
Using this relationship in the distance formula for sound allows us to write:
.
Substituting the value of d from the first equation yields:
.
Now, after some calculations, we can proceed further:
.
Finally, the value is inserted into the initial equation:

