Until a train is a safe distance from the station, it must travel at 5 m/s. Once ti

Tristan is an engineer working on a design for a new roller coaster. He wants the car to shoot out of the station very fast but

uysha [10]

Answer:

Superconducting materials can transport electrons with no resistance, and hence release no heat, sound, or other energy forms. Superconductivity occurs at a specific material's critical temperature (Tc). As temperature decreases, a superconducting material's resistance gradually decreases until it reaches critical temperature. At this point resistance drops off, often to zero, as shown in the graph at right.

Explanation:

7 0

3 years ago

Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len

Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

4 0

3 years ago

Which statement below best reflects the energy of the rock shown in the diagram?

Natali [406]

Answer:

32

Explanation:

32

5 0

3 years ago

Consider the element carbon. How many electrons can be found in a carbon atom? How many are valence electrons?

Finger [1]

The answer is B: 6; 4. Carbon has four electrons in its outermost shell, which are its valence electrons.

8 0

3 years ago

Read 2 more answers

The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta

ivanzaharov [21]

Answer:

(a) 11.3 L

(b) 10 M

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

Luminosity of Star = 11.3 L

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

Mass of Star = 10 M

3 0

4 years ago