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icang [17]
3 years ago
14

When distances were carefully measured from the Earth to globular clusters above and below the Milky Way plane (where our view o

f them is not obscured by interstellar dust and gas), their distribution was found to be
A) spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction.
B) in a relatively flat disk almost perpendicular to the plane of the Galaxy, with a relatively higher density of clusters toward its center.
C) concentrated in the plane of the Milky Way and clustered around the Sun's position, indicating that the Sun is close to the Galaxy's center.
D) uniformly distributed throughout space, with no concentration in any area of the Milky Way.
Physics
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:

A) Spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction

Explanation:

The globular clusters are present mainly in the direction of Sagittarius with the center of the system of globular cluster being measured as a spherical cluster cloud such that the center of the Milky Way can be taken as being in the Sagittarius constellation

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A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
4 years ago
What is the total capacitance in units of mF, of the two capacitors connected in series, as shown in the diagram, when C1 = 45 m
elena-14-01-66 [18.8K]

Answer:

16.7 mF

Explanation:

The total capacitance of two capacitors connected in series is given by the formula:

\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}

in our problem, we have:

C1 = 45 mF is the capacitance of the first capacitor

C2 = 26 mF is the capacitance of the second capacitor

Substituting into the equation, we find:

\frac{1}{C}=\frac{1}{45 mF}+\frac{1}{26 mF}=0.06 mF^{-1}\\C=\frac{1}{0.06 mF^{-1}}=16.7 mF

7 0
3 years ago
Read 2 more answers
The innermost satellite of jupiter orbits the planet with a radius of 422 × 103 km and a period of 1.77 days. what is the mass o
FinnZ [79.3K]

The mass of Jupitar is obtained from the calculations as 5.8 * 10^-14 Kg.

<h3>What is the mass of Jupitar?</h3>

There are nine planets in the solar system and the sun lies at the enter of our solar system. This is the heliocentric model of the solar system.

Given that;

T^2 = GMr^3/4π

T = period

G = gravitational constant

r = radius

M = mass of Jupitar

Now;

1 day = 86400 seconds

1.77 days = 1.77 days *  86400 seconds/1 day

= 152928 seconds

Making M the subject of the formula;

M =4πT^2/Gr^3

M = 4 * 3.142 * (152928)^2/6.67 × 10^-11 * (422 × 10^9)^3

M = 2.9 * 10^11/5.0 * 10^24

M = 5.8 * 10^-14 Kg

Learn more about mass of a planet:brainly.com/question/13851553

#SPJ1

5 0
2 years ago
Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1
zheka24 [161]

Answer:

v_{f}=1721.1m/s

Explanation:

Given data

q = 3.1 µC

m = 47 mg

r1 = 0.83 mm

r2 = 2.5 mm.

As we know that:

dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\    (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s

5 0
3 years ago
Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity
AfilCa [17]

The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)v

M×5+3M×0=[M+3M]v

The final velocity is found as;

V=51.25 m/s

The velocity of block A is found as;

\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\  V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s

Hence, the final velocity of the block A will be 2.5 m/sec.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ4

5 0
3 years ago
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