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Flura [38]
3 years ago
10

Listed following are several astronomical objects. Rank these objects based on their diameter, from largest to smallest. (Note t

hat the neutron star and black hole in this example have the same mass to make your comparison easier, but we generally expect black holes to have greater masses than neutron stars.)
a.main-sequence star of spectral type a.
b. jupiter.
c. a one solar mass white dwarf.
d. the moon.
e. a two solar mass neutron star.
f. event horizon of a two solar mass black hole.
Physics
1 answer:
daser333 [38]3 years ago
3 0

Answer:

Ranked

Explanation:

The rank of these objects from largest to smallest based on diameter is as follows

1. main-sequence star of a spectral type A

2. Jupiter

3. a one-solar-mass white dwarf

4.the Moon

5. a two-solar-mass neutron star

6. the event horizon of a two-solar-mass black hole

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Answer:

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Explanation:

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3 0
3 years ago
Basalt, formed from magma, contains iron-rich minerals. Why is this important?
Svetach [21]
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5 0
3 years ago
A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
valentinak56 [21]

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0
3 years ago
Water enters a house at 1.5 m/s through a pipe with an inside radius of 1.0cm and at a pressure of 400,000 Pa. The water then tr
kupik [55]

Answer:

pressure of the water = 3.3 × 10^{5} pa

Explanation:

given data

velocity v1 = 1.5 m/s

pressure P = 400,000 Pa

inside radius r1 = 1.00 cm

pipe radius r2 = 0.5 cm  

h1 = 0 (datum at inlet)

h2 = 5.0 m (datum at inlet)

density of water ρ = 1000 kg/m³

to find out

pressure of the water

solution

we consider here flow speed in bathroom that is =  v2 and Pressure in bathroom is =  P2  

here we will use both continuity and Bernoulli equations

because here we have more than one unknown so that  

v1 × A1 = v2 × A2 × P1 +  ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²

now we use here first continuity equation for get v2

v2 = v1 \frac{A1}{A2}    

v2 = 1.5 \frac{\pi * 0.01^2}{\pi * 0.005^2}    

v2 = 6 m/s

and now we use here bernoulli eqution for find here p2 that is

P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )

P2 = 400000  - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )

P2 = 3.3 × 10^{5} pa

3 0
3 years ago
If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 55 MN/m , determine the displacement of e
Alinara [238K]

Answer:

σ = 1.09 mm

Explanation:

<u>Step 1:</u> Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

<u>Step 2:</u> calculate length of the rod, L

K = \frac{A*E}{L}

L = \frac{A*E}{K}

A=\frac{\pi d^{2}}{4}

d = 20-mm = 0.02 m

A=\frac{\pi (0.02)^{2}}{4}

A = 0.0003 m²

L = \frac{A*E}{K}

L = \frac{(0.0003142)*(200X10^9)}{55X10^6}

L = 1.14 m

<u>Step 3:</u> calculate the displacement of the rod, σ

\sigma = \frac{F*L}{A*E}

\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

3 0
3 years ago
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