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2 years ago

What is the minimum mass needed for a star to be on the main sequence? What happens to stars that do not meet the minimum mass?

1 answer:

6 0

Main sequence stars are characterised by the source of their energy.They are all undergoing fusion of hydrogen into helium within their cores. The mass of the star is the main element for such process or phenomenon to take place for it is a determinant of both the rate at which they perform the said activity and the amount of fuel available.

To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.

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What do you need to know to establish motion

avanturin [10]

Answer:

An object is in motion if it changes position relative to a reference point.

Explanation:

7 0

2 years ago

A baseball rolls off a 1.20m high desk and strikes the floor 0.50m away from the base of the desk . How fast was it rolling?

noname [10]

The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration ( $g=9.8 m/s^2$ , acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

$g=9.8 m/s^2$

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s$

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

in a time

t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

$v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s$

So, the ball rolls off the table at 1.01 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

2 years ago

Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

shtirl [24]

Answer:

$L=6.21m$

Explanation:

For the simple pendulum problem we need to remember that:

$\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0$ ,

where $\theta$ is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

$\omega^{2}=\frac{g}{L}$ ,

where $\omega$ is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

$\omega=\frac{2\pi}{T}$ ,

where T is the oscillation period. Now, we can easily solve for L:

$(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m$

3 0

3 years ago

A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.

Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

$d=30,000 pc =9.26\cdot 10^{20} m$

The speed of the cosmic ray is

$v=0.98 c$

where

$c=3.0 \cdot 10^8 m/s$ is the speed of light. Substituting,

$v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s$

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

$T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s$

Converting into years,

$T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years$

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

$T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}$

And substituting v = 0.98c, we find:

$T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years$

3 0

3 years ago

In the national grid,what does the step up transformer do?

amm1812

In the National Grid, the Step Up Transformer increases the amount of voltage that is passing through the grid at any one time. These are used at power stations, as without them, there would not be enough power to transport the electricity around the national grid.
Hope this helps :)

7 0

2 years ago

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