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1 year ago

The speed of light is a constant and approximately equal to 300,000,000 meters per second.

Physics

1 answer:

sveticcg [70]1 year ago

4 0

The material used before doubling will produce light at a frequency of 0.056 Hz.

<h3 /><h3>What is the frequency?</h3>

The number of times a wave occurs repeatedly in a second is known as frequency.

The formula provides the frequency as,

$\rm v = \lambda \times f \\\\\ \rm f = \frac{v}{\lambda} \\\\ \rm f = \frac{3 \times 10^8}{532 \times 10^-9} \\\\ \rm f = 3 \times 10^{12} \\\\ f= 0.056 \ Hz$

Therefore, the material used before doubling will have an output light frequency of 0.056 Hz.

To learn more about the frequency refer to:

brainly.com/question/14926605

#SPJ1

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You are In-line skates at the top of a small hill. Your potential energy is equal to 1000 J. The last time we checked, your mass

stira [4]

A) <u>Weight = mass × acceleration (due to gravity) </u>

= 60×9.8

= 588 N

<u>B) Potential energy = mass x gravity x change in height </u>

1,000 = 60.0 x 9.8 x h

h = 1.7 m

<u>C) Kinetic energyF = potential energyI </u>

KEF = 1/2mv2

PEI = mgh = 1,000 J

1/2mv2 = 1,000

1/2(60.0)v2 = 1,000

v2 = 33.33

v = 5.77 m/s

3 0

3 years ago

In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an

natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

$A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}$

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

$=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa$

σ=5517.25 Pa

Strain in x direction

ε=σ/E

$=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}$

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0

2 years ago

Plz Help me with this

AlexFokin [52]

Answer:

Cools ; size

Explanation:

The rate at which magma cools determines the size of the crystals in the new rock. Igneous rocks are formed from the cooling and solidification of molten magma which finds its way to the surface or depth of very low pressure beneath the surface. This place or depth of cooling of magma affects the cooling rate and hence the size of the crystals formed. Igneous rocks formed at depths below the surface have more time to cool and allows more time for Crystal growth and hence produce coarse grained crystal grains called Intrusive igneous rocks which have significantly larger crystals than those formed on the surface which cools rapidly and allowing very little time for crystal growth giving rise to the formation of fine grained crystals and are called extrusive igneous rocks.

6 0

3 years ago

(01.06 MC)

Cerrena [4.2K]

Answer:

Explanation:

7 0

2 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

2 years ago

Read 2 more answers