Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
<h3>b) Time</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating :
(5)
(6)
(7)
<h3>a) Final velocity</h3>
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
<h3>c) Range</h3>
Substituting (7) in (2):
(11)
(12)
Answer:
115 kPa
Explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
Assuming no elevation change, h₁ = h₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
Plugging in values:
(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²
P = 115,000 Pa
P = 115 kPa
The density of an object can be calculated using the formula Density = Mass/Volume.
Experimental Density:
Density = 153.8g / 20.00 cm^3
Density = 7.69g/cm^3
Percent error equation:
% Error = | Theoretical Value - Experimental Value|/Theoretical Value * 100
% Error = | 7.87g/cm^3 - 7.69g/cm^3|/7.87g/cm^3 * 100
% Error = 2.29%
Therefore a is the correct answer.
