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3 years ago

What do you spiral and elliptical galaxies have in common

1 answer:

3 0

They have a uniform luminosity and are similar to the bulge in a spiral galaxy, but with no disk. The stars are old and there is no gas present. Ellipticals are usually found in the high density field, at the centre of clusters.

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How many elections are shared between one nitrogen atom and one carbon atom?

Charra [1.4K]

Answer:

3 electrons from nitrogen and 3 from carbon while carbon already has a lone pair along with a negative charge (called cyanide)

8 0

3 years ago

if the train is accelerating and the bisicle is traveling at a constant velocity, what do you know about their speed?

Vanyuwa [196]

The train is accelerating meaning there is a change in the velocity so the speed is either increasing or decreasing depending. The bicycle is travelling at a constant velocity meaning it is travelling at a constant speed.

3 0

3 years ago

What property do protons can electrons have that neutrons do not ?

Vesna [10]

Answer:

Protons and electrons are charged particles. Neutrons have no charge.

4 0

3 years ago

A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0

barxatty [35]

Answer:

Velocity = 3.25[m/s]

Explanation:

This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.

The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

$(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]$

4 0

3 years ago

A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr

Papessa [141]

Answer:

The moment of inertia about an axis through the center and perpendicular to the plane of the square is

$I_s = \frac{Ma^2}{3}$

Explanation:

From the question we are told that

The length of one side of the square is $a$

The total mass of the square is $M$

Generally the mass of one size of the square is mathematically evaluated as

$m_1 = \frac{M}{4}$

Generally the moment of inertia of one side of the square is mathematically represented as

$I_g = \frac{1}{12} * m_1 * a^2$

Generally given that $m_1 = m_2 = m_3 = m_4 = m$ it means that this moment inertia evaluated above apply to every side of the square

Now substituting for $m_1$

$I _g= \frac{1}{12} * \frac{M}{4} * a^2$

Now according to parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

$I_a = I_g + m [\frac{q}{2} ]^2$

=> $I_a = I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2$

substituting for $I_g$

=> $I_a = \frac{1}{12} * \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2$

=> $I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}$

=> $I_a = \frac{Ma^2}{12}$

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

$I_s = 4 * I_a$

=> $I_s = 4 * \frac{Ma^2}{12}$

=> $I_s = \frac{Ma^2}{3}$

8 0

3 years ago