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marin [14]
3 years ago
7

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

the object makes an angle α0 with the horizontal. Neglect air resistance. If α0 is 90∘, so that the object is thrown straight up (but misses the roof on the way down), what is the speed v of the object just before it strikes the ground?
Physics
1 answer:
Vsevolod [243]3 years ago
6 0

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

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Function of a simple pendulum​
Misha Larkins [42]

Answer:

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From a historical point of view, pendulums became important for time measurement. Simply counting the oscillations of the pendulum, or attaching the pendulum to a clockwork can help you track time. Making the pendulum in such a way that it holds its shape and dimensions (in changing temperature etc.) and using mechanisms that counteract damping due to friction led to the creation of some of the first very accurate all-weather clocks.

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:

7 0
3 years ago
Im sorry i dont understand pls helpp -w-
rodikova [14]
Temperature is related to a physical change of a substance because it determines the state of the object via cooling, heating or freezing. To make the concept clear, hear are some examples:-
1. Chocolate melts and turns into liquid form. This is a physical change
2. Water freezes via cooling and forms ice which is in a solid state. This is another physical change

Hope this helped and have a nice day : )

7 0
3 years ago
Read 2 more answers
HELP PLZ ASAP!!
Andre45 [30]
1. The successful launch of communication satellites

2.More people will be interested in studying science

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5 0
3 years ago
when 10 similar coins are dropped into a graduated cylinder from 75 ml to 100ml. what is average volume of each coin?​
Alex787 [66]

Explanation:

Volume of 10 coins = 100ml - 75ml = 25ml

Volume of 1 coin = 25ml / 10 = 2.5ml

The average volume of each coin is 2.5ml.

8 0
3 years ago
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Find deacceleration An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When
earnstyle [38]

Answer:

The deceleration is  a = -0.7273 \ m/s^2  

Explanation:

From the question we are told that

  The distance of the car from the crossing is d = 360 \  m

   The speed is u  = 16 \  m/s

    The reaction time of the engineer is  t = 0.53 \ s

Generally the distance covered during the reaction time is  

      d_r = u * t

=>   d_r = 16 * 0.53

=>   d_r = 8.48 \ m

Generally distance of the car from the crossing after the  engineer reacts is

      D = d- d_r        

=> D = 360 - 8.48      

=> D = 352 \ m

Generally from kinematic equation

      v^2 = u^2 + 2as

Here v is the final velocity of the car which is  0 m/s

So

        0^2 = 16^2 + 2 * a * 352

=>    a = -0.7273 \ m/s^2  

3 0
3 years ago
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