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CaHeK987 [17]
3 years ago
9

A projectile is fired at an angle above the level ground. Where does it have maximum kinetic energy? Is it right when it leaves

the launcher and right before it hits the ground?
Physics
2 answers:
Lorico [155]3 years ago
5 0
You can approach this problem in an analytical way. The formula for kinetic energy is 1/2*mv². Throughout the projectile motion, mass is constant. Therefore, it wholly depends on the velocity.<em> So, the maximum kinetic energy is when the object travels the fastest. This is right when it leaves the launcher, and the moment just before it hits the ground.</em>
seropon [69]3 years ago
3 0
<h2>Answer:</h2>

<u>It has maximum kinetic energy at the </u><u>starting of projectile and the end of projectile.</u>

<h2>Explanation:</h2>

If a projectile is projected at some angle to the horizontal then have maximum kinetic energy at the starting of projectile and the end of projectile. Since Ke=1/2*mv² and here the velocity is the resultant of two velocities at horizontal direction and vertical direction which is given by the parallelogram law of vector. With the passage of time due to the resistance of air friction its speed is reduced and when it again comes to the ground the speed increases due to gravitational pull.

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Answer:

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Explanation:

Let \theta denote the angle between the wire and the magnetic field.

Let B denote the magnitude of the magnetic field.

Let l denote the length of the wire.

Let I denote the current in this wire.

The magnetic force on the wire would be:

F = B \cdot l \cdot I \cdot \sin(\theta).

Because of the \sin(\theta) term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is 90^\circ.)

In this question:

  • \theta = 20^\circ (or, equivalently, (\pi / 9) radians, if the calculator is in radian mode.)
  • B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T.
  • l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m.
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Rearrange the equation F = l \cdot I \cdot \sin(\theta) to find an expression for I, the current in this wire.

\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}.

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Explanation:

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The inhaled gas contains Oxygen and majorly Nitrogen.

The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.

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