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antiseptic1488 [7]
3 years ago
10

A heated iron is pressed against a shirt, warming it. This is an

Physics
1 answer:
Nady [450]3 years ago
6 0

Answer:

Convection? I'm pretty sure that's it

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If the sound source is moving then the pitch of the sound will ___________________ to the observer.
Alika [10]

Answer:

Increase

Explanation:

The frequency of sound determines the sound. If the frequency is lower the pitch will be lower. If the frequency is higher the pitch will be higher. This is affected by the motion of the sound source because when a sound source is moving faster the frequency will become higher.

3 0
3 years ago
What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s
Furkat [3]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity v and at a distance of r from the axis of rotation is given as:

a=\frac{v^2}{r}

Here, v=8\ m/s,r=0.70\ m

∴ a=\frac{8}{0.70}=11.43\ m/s^2

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

3 0
3 years ago
Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit
cestrela7 [59]

Answer:

v=1667.9km/h

a_{cp}=436.6km/h^2

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth C=2\pi R, where R=6371km is the radius of the Earth.

We have then:

v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h

And then we use the centripetal acceleration formula:

a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2

6 0
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.

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
Which statement describes the effects of forces on an object
lukranit [14]

Answer:

Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.

Explanation:

8 0
3 years ago
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