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2 years ago

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Most interstellar clouds are: Most interstellar clouds are: much bigger than our solar system. similar in size to clouds in Eart

h's atmosphere. about the size of Earth. about the size of a wavelength of light. much smaller than the clouds in Earth's atmosphere.

1 answer:

cupoosta [38]2 years ago

4 0

Answer:

Most interstellar clouds are much bigger than our solar system.

Explanation:

An interstellar cloud refers:

It is generally an accumulation of gas, plasma, and dust in our and other galaxies.
It is basically a denser-than-average region of the interstellar medium (ISM).

Interstellar clouds can be large up to 106 solar masses

It is also often said to be the most massive entities in the galaxy.

Hence

we can say about Interstellar clouds,

They are much bigger than our solar system.

learn more about interstellar clouds here:

<u>brainly.com/question/14726563</u>

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#SPJ4

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Calculate the weight of a 1 kg mass at earth's surface. The mass of the of the Earth's surface if the mass of the earth is 6 x 1

yuradex [85]

Answer:

9.8N

Explanation:

Here we can get gravitational acceleration according to the place where object is placed by bellow equation

g = GM/R²

g - Gravitational Acceleration

G - Gravitational constant (6.67×10-11)

R - Distance ( Radius )

g = 6.67 × 10-11 × 1024 /(6.37×106)²

g = 9.8 m/s²

There for

Weight = Mass × Gravitational acceleration

= 1×9.8

= 9.8 N

4 0

3 years ago

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

What is the magnitude of the acceleration vector which causes a particle to move from velocity −5i−2j m/s to −6i+ 7j m/s in 8 se

shusha [124]

Answer:

Acceleration, $a=\dfrac{1}{8}(-i+9j)\ m/s^2$

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

Final velocity of particle in vector form, v = (-6i + 7j) m/s

Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}$

$a=\dfrac{(-i+9j)}{8\ s}\ m/s^2$

or

$a=\dfrac{1}{8}(-i+9j)\ m/s^2$

Hence, the value of acceleration vector is solved.

4 0

3 years ago

If a compound has a very low melting and boiling point, it is likely that the compound possesses mainly which type of intermolec

sergejj [24]

For the given question above, I think there is an associated choice of answer for it. However, the answer for this is London Dispersion Forces. <span>Dipole-dipole forces and hydrogen bonding are much stronger, leading to higher melting and boiling points.</span>

6 0

4 years ago

What current flows through a 2.56-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 ✕ 103 V is applied to it? (Suc

vfiekz [6]

Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, $\rho=2300\ \Omega-m$

Length of rod, l = 20 cm = 0.2 m

Voltage, $V=1\times 10^3\ V$

The resistivity of the rod is given by :

$R=\rho\dfrac{L}{A}$

$R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}$

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

$I=\dfrac{V}{R}$

$I=\dfrac{10^3\ V}{893692.30\ \Omega}$

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

6 0

3 years ago