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telo118 [61]
3 years ago
15

Group 2 and Group 3 stars represent which stages of development? Describe these groups in terms of their relative age, size, bri

ghtness, and temperature.
Physics
1 answer:
gladu [14]3 years ago
8 0

Answer:

it represents group 2 and also it represents the group 3 stars

Explanation:

because i just took it and got it correct.

You might be interested in
g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for
lbvjy [14]

Answer:

2

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

a\text{sin}\theta=\frac{2N+1}{2}\lambda

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}

Solving the above equation we get

we N= 2.313≅ 2

4 0
3 years ago
What new tool did geologists use to estimate the changing flow rates of lava over the duration of this eruption? Hint: The answe
Vladimir [108]

Answer:

Option b, pothographs from drones.

Explanation:

the USGS (U.S. Geological Survey) decided to make photographic captures from drones to the volcanic surfaces, which allowed through observations to understand things like the characteristics of the lava, the height of the volcanic plumes (among others).

Podemos ver en el siguiente enlace un ejemplo de fotografía tomada desde un dron al Kilauea.

https://www.usgs.gov/media/images/k-lauea-volcano-drone-over-lava-channel

4 0
3 years ago
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
4 years ago
for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole
sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

Cp = 5R/2

3 0
3 years ago
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
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