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Marina86 [1]
3 years ago
8

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

te a number.
Physics
1 answer:
pickupchik [31]3 years ago
4 0
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
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Murrr4er [49]

Answer:

c

Explanation:

i would think c would be correct because

blood temperature IS affected by muscles

digestion IS affected by muscles

speech IS affected by muscles

so therefore <em>hair growth</em> IS NOT affected by muscles

<h2><u><em>Please mark Brainliest</em></u></h2>
3 0
3 years ago
A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.
myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

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3 years ago
Function of a simple pendulum​
Misha Larkins [42]

Answer:

A pendulum is a mechanical machine that creates a repeating, oscillating motion. A pendulum of fixed length and mass (neglecting loss mechanisms like friction and assuming only small angles of oscillation) has a single, constant frequency. This can be useful for a great many things.

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Pendulums were/are also important for musicians, where mechanical metronomes are used to provide a notion of rhythm by clicking at a set frequency.

The Foucault pendulum demonstrated that the Earth is, indeed, spinning around its axis. It is a pendulum that is free to swing in any planar angle. The initial swing impacts an angular momentum in a given angle to the pendulum. Due to the conservation of angular momentum, even though the Earth is spinning underneath the pendulum during the day-night cycle, the pendulum will keep its original plane of oscillation. For us, observers on Earth, it will appear that the plane of oscillation of the pendulum slowly revolves during the day.

Apart from that, in physics a pendulum is one of the most, if not the most important physical system. The reason is this - a mathematical pendulum, when swung under small angles, can be reasonably well approximated by a harmonic oscillator. A harmonic oscillator is a physical system with a returning force present that scales linearly with the displacement. Or, in other words, it is a physical system that exhibits a parabolic potential energy.

A physical system will always try to minimize its potential energy (you can accept this as a definition, or think about it and arrive at the same conclusion). So, in the low-energy world around us, nearly everything is very close to the local minimum of the potential energy. Given any shape of the potential energy ‘landscape’, close to the minima we can use Taylor expansion to approximate the real potential energy by a sum of polynomial functions or powers of the displacement. The 0th power of anything is a constant and due to the free choice of zero point energy it doesn’t affect the physical evolution of the system. The 1st power term is, near the minimum, zero from definition. Imagine a marble in a bowl. It doesn’t matter if the bowl is on the ground or on the table, or even on top of a building (0th term of the Taylor expansion is irrelevant). The 1st order term corresponds to a slanted plane. The bottom of the bowl is symmetric, though. If you could find a slanted plane at the bottom of the bowl that would approximate the shape of the bowl well, then simply moving in the direction of the slanted plane down would lead you even deeper, which would mean that the true bottom of the bowl is in that direction, which is a contradiction since we started at the bottom of the bowl already. In other words, in the vicinity of the minimum we can set the linear, 1st order term to be equal to zero. The next term in the expansion is the 2nd order or harmonic term, a quadratic polynomial. This is the harmonic potential. Every higher term will be smaller than this quadratic term, since we are very close to the minimum and thus the displacement is a small number and taking increasingly higher powers of a small number leads to an even smaller number.

This means that most of the physical phenomena around us can be, reasonable well, described by using the same approach as is needed to describe a pendulum! And if this is not enough, we simply need to look at the next term in the expansion of the potential of a pendulum and use that! That’s why each and every physics students solves dozens of variations of pendulums, oscillators, oscillating circuits, vibrating strings, quantum harmonic oscillators, etc.; and why most of undergraduate physics revolves in one way or another around pendulums.

Explanation:

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3 years ago
Two spaceships leave Earth in opposite directions, each with a speed of 0.60c with respect to Earth. (a) What is the SPEED of sp
Studentka2010 [4]

Answer:

The relative speed of 1 relative to 2 is 0.88c

Explanation:

In relativistic mechanics the relative speed between 2 objects moving in different direction is given by

v_{ab}=\frac{v_{a}+v_{b}}{1+\frac{v_{a}v_{b}}{c^{2}}}

Since it is given that

v_{a}=0.6c\\\\v_{b}=0.6c

Applying values in the formula we get

v_{ab}=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}}\\\\v_{ab}=0.88c

8 0
4 years ago
A hockey player skates with an average speed of 8.25 m/s. The distance the player will travel in 5.60 s is __________ .
OverLord2011 [107]
<h2>Greetings!</h2>

To find this value, you need to remember the speed formula:

3 = 6 / 2

Speed = distance ÷ time

Rearrange to make distance the subject:

Distance = speed * time

Simply plug these values into this:

5.6 * 8.25 = 46.2

<h3>So the player will travel 46.2 metres!</h3>
<h2>Hope this helps!</h2>
7 0
3 years ago
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