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Strike441 [17]
2 years ago
3

Earth's tectonic activity is a significant factor that made life on Earth possible because the movement of Earth's crustal plate

s affects _____.
Physics
1 answer:
Genrish500 [490]2 years ago
8 0
Earth's tectonic activity is a significant factor that made life on Earth possible because the movement of Earth's crustal plates affects volcanic activity.

<span>♡♡Hope I helped!!! :)♡♡
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The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61
nikitadnepr [17]

Answer:

(a) Angle of incidence for violet is more than the angle of incidence for red

(b) 2.4°

Explanation:

refractive index for violet , v = 1.66

refractive index for red, nR = 1.61

wavelength for violet, λv = 400 nm

wavelength for red, λR = 700 nm

Angle of refraction, r = 30°

(a) Let iv be the angle of incidence for violet.

Use Snell,s law

nv = Sin iv / Sin r

1.66 = Sin iv / Sin 30

Sin iv = 0.83

iv = 56°

Use Snell's law for red

nR  = Sin iR / Sin r  

where, iR be the angle of incidence for red

1.61 = Sin iR / Sin 30

Sin iR = 0.805

iR = 53.6°

So, the angle of incidence for violet is more than red.

(b) iv - iR = 56° - 53.6° = 2.4°

4 0
2 years ago
In the metric system, the appropriate unit for weight is the _____. gram newton newton/cm2 gram/cm3
Archy [21]

Answer:

Newton

Explanation:

The earth attracts every body towards its centre. The force with which the earth attracts any body towards its centre, is called its weight.

It is a vector quantity.

It always acts towards the centre of earth.

The SI unit of Newton.

4 0
2 years ago
A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?
omeli [17]
The ball hitting the bat
6 0
2 years ago
Read 2 more answers
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
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calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it
Lera25 [3.4K]
The force of F=10 N produces an extension of
x=4 cm=0.04 m
on the string, so the spring constant is equal to
k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m

Then the string is stretched by \Delta x=40 cm=0.40 m. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
W= \Delta U= \frac{1}{2}k(\Delta x)^2  = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J
5 0
2 years ago
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