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3 years ago

Which is likely to be more common in our Galaxy: white dwarfs or black holes? Why?

Physics

1 answer:

Nookie1986 [14]3 years ago

4 0

Answer:

White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.

You might be interested in

How come the arrow points the other direction when we look at it through a glass of water?

melomori [17]

Reflection changes the way the light hits our eyes, changing the direction.

5 0

3 years ago

In a microwave oven, electrons describe circular motion in a magnetic field within a special tube called amagnetron; as you'll l

QveST [7]

Answer:

The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

Explanation:

Given that,

Diameter = 2.62 mm

Frequency = 2.15 GHz

(A). We need to calculate the magnetic field strength

Using formula of the magnetic field strength

$B=\dfrac{2\pi mf}{e}$

Where, f = frequency

e = charge of electron

Put the value into the formula

$B=\dfrac{2\times3.14\times9.1\times10^{-31}\times2.15\times10^{9}}{1.6\times10^{-19}}$

$B=0.077\ T$

(B). We need to calculate the energy of electron

Using formula of energy

$E=\dfrac{1}{2}m(r\omega)^2$

$E=\dfrac{1}{2}\times9.1\times10^{-31}\times(1.31\times10^{-3}\times2\pi\times2.15\times10^{9})^2$

$E=1.4249\times10^{-16}\ J$

The energy in eV

$1 eV=1.6\times10^{-16}\ J$

$E=\dfrac{1.4249\times10^{-16}}{1.6\times10^{-16}}$

$E=0.8906\ eV$

Hence, The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

5 0

3 years ago

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given that,

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We need to calculate the value of $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, put the all values into the acceleration 's formula

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

7 0

3 years ago

Read 2 more answers

Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo

zubka84 [21]

Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

ω = 10800πrad / 60 s

ω = 180πrad / seg

ω = 565.56 rad / seg

SI)

Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

C) Periodo (T)

Recordar T = 1 / f

Por lo tanto,

T = 1/90

T = 0.0111111s

3 0

3 years ago

a rock is dropped from a height of 80 m and is in free fall what is the velocity of as it reaches the ground 4.0 seconds later

Marta_Voda [28]

Velocity = displacement (distance)/time

v=80m/4s

v=20m/s

velocity = 20 meters per second

8 0

3 years ago