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3 years ago

How are stars distributed throughout an elliptical galaxy?

1 answer:

6 0

An elliptical galaxy is a type of galaxy having an approximately ellipsoidal shape and a smooth, nearly featureless brightness profile. Unlike flat spiral galaxies with organization and structure, they are more three-dimensional, without much structure, and their stars are in somewhat random orbits around the center.

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Physics Homework MathPhys homie if you see this pls help

cluponka [151]

Answer:

1. -8.20 m/s²

2. 73.4 m

3. 19.4 m

Explanation:

1. Apply Newton's second law to the car in the y direction.

∑F = ma

N − mg = 0

N = mg

Apply Newton's second law to the car in the x direction.

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Given μ = 0.837:

a = -(9.8 m/s²) (0.837)

a = -8.20 m/s²

2. Given:

v₀ = 34.7 m/s

v = 0 m/s

a = -8.20 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx

Δx = 73.4 m

3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.

d = v₀t

d = (34.7 m/s) (0.56 s)

d = 19.4 m

6 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

The electric potential at the surface of a charged conductor _______.

frez [133]

Answer:

The electric potential at the surface of a charged conductor is always such that the potential is zero at all points inside the conductor.

Explanation:

Each point on the surface of a balanced charged conductor has the same electrical potential.

The surface on any charged conductor in electrostatic equilibrium is an equipotential surface. Since the electric field is equal to zero inside the conductor, the electric potential at any point inside and on the surface is equivalent to its value.

6 0

3 years ago

Which of the following statements is accurate?

cupoosta [38]

Answer: A and B

Explanation:

The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Because wavelength is the distance between the two successful crest or trough.

Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.

Because amplitude is the measure of maximum displacement from the original position

4 0

3 years ago

How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.

pav-90 [236]

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

8 0

3 years ago