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Tatiana [17]
3 years ago
15

A neutron star is the remnant left after certain supernovae (explosions of giant stars). Typically, neutron stars are about 18 k

m in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm^3?
Physics
1 answer:
Tresset [83]3 years ago
5 0

Answer:

Density=6.51*10^{14}g/cm^{3

Explanation:

Sun mass:

Ms=1.989 × 10^30 kg

Neutron star has the same mass.

Radius Neutron:

R=9Km (because diameter is 18Km)

R=9*10^3m

Density neutron star:

D=Mn/Vol=Ms/(4/3*\pi*R^3)=1.989*10^{30} /(4/3*\pi*(9*10^{3})^{3})=6.51*10^{17}kg/m^{3}

D=6.51*10^{17}kg/m^{3}*(1000g/1Kg)*(1m^{3}/1000000cm^{3})=6.51*10^{14}g/cm^{3

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An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around whic
Cloud [144]

Answer:

The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

Given that,

Length = 19 cm

Diameter = 23 cm

Time = 10 sec

Initial angular velocity = 65 rpm

Final velocity = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Using formula of tangential acceleration

a_{t}=r\alpha

a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}

a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}

a_{t}=0.0301\ m/s^2

Hence, The tangential acceleration of the pedal is 0.0301 m/s².

4 0
3 years ago
When does a magnet induce an electric current in a wire coil?
zhuklara [117]

Answer:

B I believe

Explanation:

8 0
2 years ago
If cats always land on their feet, how do they get hurt when they fall?
Ulleksa [173]
The impact front the fall, depending on the distance, has an effect on the impact of their bones crushing with the pressure from the fall.
6 0
3 years ago
A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t
Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

d = d_1 + d_2

given that

d_1 = 200 cos20 i + 200 sin20 j

d_1 = 187.94 i + 68.4 j

d_2 = -230 sin30 i + 230 cos30 j

d_2 = -115 i + 199.2 j

now the total distance is

d = (187.94 - 115)i + (199.2 + 68.4)j

d = 72.94 i + 267.6 j

now the magnitude of the distance is given as

d = \sqrt{72.94^2 + 267.6^2}

d = 277.4

also the direction is given as

\theta = tan^{-1}\frac{267.6}{72.94}

\theta = 74.7 degree

<em>so it is 277.4 km at 74.7 degree North of East</em>

8 0
3 years ago
A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?
seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

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L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0
2 years ago
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