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3 years ago

The Milky Way is a _____ galaxy.

Physics

2 answers:

Serjik [45]3 years ago

7 0

It is a barred spiral galaxy.

gogolik [260]3 years ago

3 0

Well it was just recently named by astronomers as a barred spiral galaxy

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Which process is a physical change?

zlopas [31]

Water boiling, no cheamical bonds have been altered.

7 0

3 years ago

In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the e

Mariulka [41]

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

F = m a

G m M / x² = m dv / dt = m dv/dx dx/dt

G M / x² = dv/dx v

GM dx / x² = v dv

We integrate

v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s

½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)

72.25 10⁶ - v₀² = -1.213 10⁸

v₀² = 72.25 10⁶ + 1,213 10⁸

v₀² = 193.6 10⁶

v₀ = 13.9 10³ m / s

6 0

4 years ago

A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The

Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is $W=2430\ pound$

mass of roller coaster $m=\frac{W}{g}=\frac{2430}{32.2}=75.45$

Distance traveled by roller coaster $d=396\ ft$

drag force $f_d=85\ pounds$

velocity at top $v=59 mph\approx 86.53\ ft/s$

Suppose E is the initial energy

Conserving Energy at bottom and top

$E=\frac{1}{2}mv^2+mgh+f_d\cdot d$

$E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396$

$E=2.9\times 10^5\ foot-pound$

5 0

3 years ago

Donner le lien entre la consommation d,eau et l,aspect de votre peau c,est un dm note reponse

olasank [31]

Answer:

tu es sympa

Explanation:

4 0

2 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$