Cholesterol is found in grains. true false
2 answers:
You might be interested in
The initial velocity is
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.![\frac{dv}{dt} =10-0.8v \\\\ \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \, \frac{dv}{10-0.8v} = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\ \frac{0.8v -10}{3.2} =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D10-0.8v%20%5C%5C%5C%5C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%3Ddt%20%5C%5C%5C%5C%20%5Cint_%7B16.5%7D%5E%7Bv%7D%20%5C%2C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%20%20%3D%20%5Cint_%7B0%7D%5E%7Bt%7D%20dt%20%5C%5C%5C%5C%20-%20%5Cfrac%7B1%7D%7B0.8%7D%20%5Bln%2810-0.8v%29%5D_%7B16.5%7D%5E%7Bv%7D%3Dt%20%5C%5C%5C%5C%20ln%20%5Cfrac%7B10-0.8v%7D%7B-3.2%7D%3D-0.8t%20%5C%5C%5C%5C%20%20%5Cfrac%7B0.8v%20-10%7D%7B3.2%7D%20%20%3De%5E%7B-0.8t%7D%20%5C%5C%5C%5C%200.8v%20%3D%2010%20%2B%203.2e%5E%7B-0.8t%7D%20%5C%5C%5C%5C%20v%3D12.5%2B4e%5E%7B-0.08t%7D)
The velocity function is
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
The depth of the lake is
![d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft](https://tex.z-dn.net/?f=d%3D%5Cint_%7B0%7D%5E%7B5.7%7D%20%5C%2C%20%2812.5%2B4e%5E%7B-0.8t%7D%29dt%20%5C%5C%5C%5C%20%3D%2012.5%285.7%29%2B%20%5Cfrac%7B4%7D%7B%28-0.8%29%7D%5Be%5E%7B-0.8t%7D%5D_%7B0%7D%5E%7B5.7%7D%20%5C%5C%5C%5C%20%3D71.25-5%280.0105-1%29%20%3D76.198%20%5C%2C%20ft)
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.
The velocity function is
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
The depth of the lake is
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
The motion of the package can be described as the motion of a projectile,
given that it has an horizontal velocity and it is acted on by gravity.
- a)
= 70·i
- b) The package will reach ground in approximately <u>12.77 seconds</u>.
- c) The speed of the package as it lands is approximately <u>145.51 m/s</u>.
- d) The path of the package based on a stationary frame of reference is <u>parabolic</u>
- e) The path of the package as seen from the plane is <u>directly vertical</u> downwards
Reasons:
Velocity of the aircraft = 70 m/s
Direction of flight of the aircraft = Eastward
Height from which the aircraft drops the package, h = 800 m
a) The initial velocity of the package, = 70·i
b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
Which gives;
The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>
c) The vertical velocity just before the package reaches the ground, , is given as follows;
= 2·g·h
Therefore;
= √(2·g·h)
Which gives;
= √(2 × 9.81 × 800) ≈ 125.28
≈ 125.28 m/s
Which gives; = 70·i - 125.28·j
Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51
The speed of the package as it lands, |v| ≈ <u>143.51 m/s</u>
d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.
Therefore;
The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.
e) As seen by someone on the aeroplane, the horizontal velocity will be
zero, therefore, the package will appear as accelerating <u>directly vertical</u>
downwards.
Learn more about projectile motion here: