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Fiesta28 [93]
3 years ago
14

What is the difference between a do universal law and a scientific theory

Physics
1 answer:
liubo4ka [24]3 years ago
6 0

The main difference between a law and a theory is that a theory tries to explain the reasoning behind something that occurs in nature, whereas scientific laws are just descriptive accounts of how something occurs in nature. Law and theory are two terms that are often used in context of scientific terminology.

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What is the oxidation state of beryllium
Paha777 [63]
<span> Beryllium has an exclusive </span>+2<span> oxidation state in all of its compounds</span>
3 0
3 years ago
A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
Oksi-84 [34.3K]
<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

3 0
3 years ago
If the mass of one of two objects is increased, the force of attraction between them will
Sonja [21]
It will increase..........................
5 0
3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

hence it is a subsonic aircraft

4 0
3 years ago
When the displacement of a mass on a spring is 12 a the half of the amplitude, what fraction of the mechanical energy is kinetic
Sveta_85 [38]
You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).

And you know that the potential energy, PE, is [ 1/2 ] k (x^2)

Then, use x = A, to calculate the PE in the point where ME = PE.

ME = PE = [1/2] k (A)^2.

At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2

=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME

So, if PE is 1/4 of ME, KE is 3/4 of ME.

And the answer is 3/4


7 0
3 years ago
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