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3 years ago

What is the frequency of a wave having a period equal to 18 seconds

Physics

1 answer:

Ivanshal [37]3 years ago

3 0

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

The relation between wave period and frequency is as follows.

T = \frac{1}{f}T=

where, T = time period

f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

T = \frac{1}{f}T=

or, f = \frac{1}{T}f=

= \frac{1}{18 sec}

18sec

= 0.055 per second (1cycle per second = 1 Hertz)

or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz

<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>

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A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

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$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

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3 years ago

Steel Usually forms a

meriva

Answer:

Permanent magnetism (of the steel)

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3 years ago

A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/

larisa [96]

Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, $m_1=200\ g= 0.2\ kg$

Mass of the second piece, $m_2=300\ g= 0.3\ kg$

Speed of the first piece, $v_1=82\ m/s$ (along x axis)

Speed of the second piece, $v_2=45\ m/s$ (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

$p_x=m_1v_1$

$p_x=0.2\ kg\times 82\ m/s$

$p_x=16.4\ kg-m/s$

$p_y=m_2v_2$

$p_y=0.3\ kg\times 45\ m/s$

$p_y=13.5\ kg-m/s$

The net momentum is given by :

$p=\sqrt{p_x^2+p_y^2}$

$p=\sqrt{16.4^2+13.5^2}$

p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

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2 years ago

Estimate the mass of blood in your body Explain your reasoning (Note: It is not enough to provide a numeric answer. The main poi

AVprozaik [17]

Answer: A little more that 5 Kg for a healthy person

Explanation: First, we know the following:

The regular adult has from 9 to 12 pints of blood. This is around 5 liters for a healthy male adult.

The human body is composed mostly on water, around 80%.

Blood is mostly composed on plasma, which makes blood thicker than water.

Knowing that, almost all the body is compose of water, it is safe to think that blood density should be near to that of water but higher.

The density on water is a know value. Which makes the following true:

<em>1 Liter of Water weights 1 Kg</em>

<em />

It could be said then, that the total mass of blood for a healthy person should be a little more that 5 kgs.

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2 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

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3 years ago

What is the frequency of a wave having a period equal to 18 seconds ​

What is the frequency of a wave having a period equal to 18 seconds