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3 years ago

The speed of a wave changes based on _ &

Physics

1 answer:

Yuki888 [10]3 years ago

4 0

Frequency and Wavelength

<u>Explanation:</u>

The speed of a wave changes based on frequency and wavelength. Wavelength is the distance between two corresponding points on adjacent waves. Wave frequency is the number of waves that pass a fixed point in a given amount of time. The wave speed depends upon the medium through which the wave is moving. Only an alteration in the properties of the medium will cause a change in the speed.

Speed, frequency and wavelength is related as:

speed = frequency X wavelength

Increasing the wavelength of a wave doesn’t change its speed. That’s because when wavelength increases, wave frequency decreases. As a result, the product of wavelength and wave frequency is still the same speed.

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What is one way you can state how you feel?

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Words and more words

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3 years ago

A student stands with both feet on some scales in order to measure his weight.

Neko [114]

Answer:

500 N

Explanation:

Because even if he lift one foot his weight will be same as the pressure applied on the scale will be the same and will not change it is not that the scale measures each foot separately

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3 years ago

Mazie stands on her kitchen floor. The coefficient of kinetic friction between her socks and the floor is 0.35, and the coeffici

Natasha2012 [34]

Static friction is a force that keeps an object at rest. Static friction definition can be written as: The friction experienced when individuals try to move a stationary object on a surface, without actually triggering any relative motion between the body and the surface on which it is on

<h3>How can calculate the Fs by finding your normal force and multiplying it by your static friction force?</h3>

N = mg = (9.8)×(58) = 568.4N

Fs = NKs = (568.4)×(0.42) =238.7N

b) So this is an interesting question, she's going at a speed of 0.6 m/s. Kinetic friction is pushing her back so theres a negative force which is the force of kinetic friction right. So its NKf = (238.7)×(0.35) = 83.5N

To learn more about Static friction, refer

brainly.com/question/13680415

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2 years ago

In which position does the ball have the least potential energy but the most kinetic energy

Firdavs [7]

Answer:

position 2

Explanation:

the gravity from the ball to the floor had the most movement involved.

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4 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

The speed of a wave changes based on _________________________ & ________________________

The speed of a wave changes based on _ &