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3 years ago

How is a scientific law alike as a societal law

1 answer:

5 0

Describe that a scientific law is a description of a specific relationship under given conditions. Describe that scientific laws are developed from large amounts of experimental observations that result in the same outcome.

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What is the speed of sound at sea level?

svetlana [45]

The speed of sound at sea level is 340.29 m/s (meters per seconds).

7 0

3 years ago

Whats the force of gravitation of a 10kg rock and 100kg boulder which are 5 meters apart

spayn [35]

Answer:

F = 2.6692 x 10⁻⁹ N

Explanation:

Given,

The mass of the rock, m = 10 kg

The mass of the boulder, M = 100 kg

The distance between them, d = 5 m

The gravitational force between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by the formula

<em> F = GMm/d² newton</em>

Where,

G - Universal gravitational constant

Substituting the given values,

F = 6.673 x 10⁻¹¹ x 100 x 10 / 5²

F = 2.6692 X 10⁻⁹ N

Hence, the force between the two bodies is, F = 2.6692 X 10⁻⁹ N

6 0

3 years ago

PLEASE HURRY!

zloy xaker [14]

PRETTY SURE ITS A AND C BUT I LOOKED IT UP AND IM ALMOST POSITIVE THARS IT

8 0

3 years ago

Read 2 more answers

The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency? Hz(b) What is its wavelength in glas

vekshin1

<h2>Answers:</h2>

The speed of a wave is given by:

$v=f.\lambda$ (1)

Where $f$ is the frequency and $\lambda$ the wavelength.

In the case of light, its speed is:

$c=f.\lambda$ (2)

On the other hand, the described situation is known as Refraction, a phenomenon in which the light changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (3)

In addition, as the light changes its direction, its wavelength changes as well:

$n=\frac{\lambda_{air}}{\lambda_{glass}}$ (4)

Knowing this, let's begin with the answers:

<h2>a) Frequency</h2>

From equation (2) we can find $f$ :

$f=\frac{c}{\lambda}$ (5)

Knowing that $1nm=(10)^{-9}m$ :

$f=\frac{3(10)^{8}m/s}{632.8(10)^{-9}m}$

$f=4.74(10)^{14}Hz}$ (6) >>>Frequency of the helium-neon laser light

<h2>b) Wavelength in glass</h2>

We already know the wavelength of the light in air $\lambda_{air}$ and the index of refraction of the glass.

So, we only have to find the wavelength in glass $\lambda_{glass}$ from equation (4):

$\lambda_{glass}=\frac{\lambda_{air}}{n}$

$\lambda_{glass}=\frac{632.8(10)^{-9}m}{1.48}$

$\lambda_{glass}=427(10)^{-9}m=427nm$ (7) >>>Wavelength of the helium-neon laser light in glass

<h2>c) Speed in glass</h2>

From equation (3) we can find the speed $v$ of this light in glass:

$v=\frac{c}{n}$

$v=\frac{3(10)^{8}m/s}{1.48}$

$v=2.027(10)^{8}m/s$ (8) >>>Speed of the helium-neon laser light in glass

7 0

2 years ago

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

Explanation:

Given that,

Electric field $E=-6.00\times10^{5}i\ N/C$

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

$F=F_{e}$

$ma=qE$

$a=\dfrac{Eq}{m}$

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

$a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}$

$a=-5.74\times10^{13}\ m/s62$

The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). We need to calculate the initial peed

Using equation of motion

$v^2-u^2=2as$

Where, s = distance

Put the value into the formula

$0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}$

$u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}$

$u=2834783.94$

$u=2.83\times10^{6}\ m/s$

The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

$t=\dfrac{u}{a}$

Where, u = initial velocity

a = acceleration

Put the value into the formula

$t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}$

$t=0.493\times10^{-7}\ sec$

Hence, (a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

5 0

2 years ago