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3 years ago

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Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of

electromagnetic waves are these? (c) How much energy (in electron volts) is carried by one quantum of this radiation?

1 answer:

laiz [17]3 years ago

8 0

Answer:

a) the frequency is 3.191x10^11

b)the radiation is a microwave radiation

c) the energy of one quantum of this radiation is 1.32x10^-3ev

Explanation:

Detailed explanation and calculation is shown in the image below

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An object that completes 20 vibrations in 10 seconds has a frequency of

nika2105 [10]

Answer:

<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>

Explanation:

<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = $\frac{20}{10}$ = 2Hz

The <em>time period</em> is defined as time taken by the object to complete one full oscillation.

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

T= $\frac{1}{2}$ =0.5 s

<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>

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3 years ago

A person on a bike (m=90kg) is traveling 4m/s at the top of a 2m hill. What is his/her

alekssr [168]

Answer:

Ug = 1764J

Explanation:

Ug = mgh

Ug = 90*9.8*2

Ug = 1764J

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2 years ago

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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

A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we

Harlamova29_29 [7]

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

$\Delta P = mgh$

$\Delta P = (8.7)(9.8)(3)$

$\Delta P = 255.78J$

Therefore the change in the internal energy of the system is 255.78

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3 years ago

A car can accelerate from rest to 28 m/s and travels 280meters. How long does this acceleration take?

Reika [66]

Answer:

$a=1.4m/s^2$

Explanation:

From the question we are told that

Velocity v=28m/s

Distance d=280

Generally the Newtons equation for motion is mathematically given as

Acceleration

$v^2=2as$

$a=\frac{V^2}{2s}$

$a=\frac{28^2}{2*280}$

$a=1.4m/s^2$

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3 years ago