All categories

3 years ago

What are the similarities and differences between novae and supernovae?

Physics

1 answer:

DochEvi [55]3 years ago

3 0

Answer:

Explanation:

A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.

A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.

From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae

You might be interested in

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

We will use the second equation of motion to find out the initial speed of the arrow.

$h= v_it + \frac{1}{2}gt^2\\$

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}$

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0

2 years ago

Which graph accurately shows the relationship between kinetic energy and speed as speed increases?

mixer [17]

Answer:

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = $\frac{1}{2}$ m $v^{2}$

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE = $\frac{1}{2}$ x 2 x $3^{2}$

= 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE = $\frac{1}{2}$ x 2 x $4^{2}$

= 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 0

3 years ago

What's the best place to watch videos for general physics?

stepan [7]

Check Khan Academy, they have explanations,notes,videos,quizzes and much more

6 0

3 years ago

Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3

cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

$J = - D \frac{\Delta c}{\Delta x}$

Where $\frac{\Delta c}{\Delta x}$ is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

$D= D_0 e^{\frac{E_d}{RT}}$

Where $D_0$ is the proportionality constant,

R is the gas constant, T the temperature and $E_d$ is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

$J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}$

Rearrange the equation to get the value of temperature,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

We have all the values in our equation.

$\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}$

$\Delta x = 9.7*10^{-3}m$

$E_d = 82000J$

$D_0 = 6.5*10^{-7}m^2/s$

$J = 3.2*10^{-9}m^2/s$

$R= 8.31Jmol^{-1}K$

Substituting,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

$T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}$

$T=1118.07K=844\°C$

4 0

4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

IgorLugansk [536]

Answer:

photo one- refraction

photo two- diffraction

photo three- reflection

8 0

3 years ago