All categories

3 years ago

On what bases are galaxies classified?

Physics

1 answer:

andreev551 [17]3 years ago

7 0

Based on the fact that the galaxy classification names all have to do with shape, your answer is C.

You might be interested in

A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.

inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

Explanation:

Given that,

Diameter = 6.40 cm

Rate of flow = 40.0 L/s

Pressure $P=1.62\times10^{6}\ N/m^2$

We need to calculate the Reynolds numbers for flow in the fire hose

Using formula of rate of flow

$Q=Av$

$v=\dfrac{Q}{A}$

Where, Q = rate of flow

A = area of cross section

Put the value into the formula

$v=\dfrac{40.0\times10^{-3}}{3.14\times(3.2\times10^{-2})^2}$

$v=12.44\ m/s$

We need to calculate the Reynolds number

Using formula of the Reynolds number

$n_{R}=\dfrac{2\rho\times v\times r}{\eta}$

Where, $\eta$ =viscosity of fluid

$\rho$ =density of fluid

Put the value into the formula

$n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}$

$n_{R}=7.945\times10^{5}$

Hence, The Reynolds numbers for flow in the fire hose.

3 0

3 years ago

If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot

Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=I_{1}+I_{2}$

$I=(m+m_{1})\times r^2+(m+m_{2})\times r^2$

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

$I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2$

$I=117.45\ kg-m^2$

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0

4 years ago

Define principal focus of concave mirror.Where should we place a candle in front of a concave mirror to get an enlarged,erect im

Mars2501 [29]

Answer

The image formed by a concave mirror will be virtual, erect and magnified when the object is placed only between the pole and the principal focus of the mirror.

Hope it helped u if yes mark me brainliest!

Tysm!

3 0

4 years ago

A basketball is resting on the ground.

kakasveta [241]

Answer:

Explanation:

there is no net force on the basketball

8 0

3 years ago

Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it

lesya692 [45]

Answer:

$v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)$

Explanation:

As we know that the mass is revolving with constant angular speed in the circle of radius R

So we will have

$\theta = \omega t$

now the position vector at a given time is

$r = Rcos\theta \hat i + R sin\theta \hat j$

now the linear velocity is given as

$v = \frac{dr}{dt}$

$v = (-R sin\theta \hat i + R cos\theta \hat j)\frac{d\theta}{dt}$

$v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)$

6 0

3 years ago