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Virty [35]
3 years ago
7

As a rubber band and moves forward, which of the following is true

Physics
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

It can go back to it's original shape

Explanation:

You might be interested in
infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on earth
Mashcka [7]

The wavelength of the infrared radiation is  λ = 3.174×10^{-5}m.

<h3>What is infrared radiation?</h3>

An infrared telescope is tuned to detect infrared radiation with a frequency of 9.45 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 9.45 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

λ=c/f

λ = 3×10^{8}/9.45×10^{12}

λ = 3.174 ×10^{-5} m

The term "infrared radiation" (IR) refers to a part of the electromagnetic radiation spectrum with wavelengths between about 700 nanometers (nm) and one millimeter (mm). Longer than visible light waves but shorter than radio waves are infrared waves.

Electromagnetic radiation with wavelengths longer than those of visible light is known as infrared, also known as infrared light. Since it is undetectable to the human eye, The typical range of wavelengths considered to be infrared (IR) is from about 1 millimeter to the nominal red edge of the visible spectrum, or about 700 nanometers.

To learn more about infrared radiation from the given link:

brainly.com/question/13163856

#SPJ4

5 0
1 year ago
Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through
Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0
2 years ago
Which of the following is a chemical change?
klemol [59]

a). Water is still H₂O after it freezes.

b). Ice is still H₂O after it melts.

c). Wire is still Cu when it's bent.

d). Paper combines with the O₂ in the air, and turns into
     a lot of new compounds when it burns.

3 0
3 years ago
How does medium affect the level/amplitude of sound?
Romashka-Z-Leto [24]
Stop drunk drive fast go
7 0
3 years ago
Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing
Juliette [100K]

Answer: MR²

is the the moment of inertia  of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements  are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

3 0
3 years ago
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