An ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any
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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:
Answer:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
Explanation:
The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>
We have a block-spring system, whose angular frequency () is defined by the following formula:
(1)
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
And the period (), measured in seconds, is determined by the following expression:
(2)
By applying (1) in (2), we get the following formula:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.