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anyanavicka [17]
3 years ago
10

Observations of the Crab Nebula taken over several decades show that gas blobs that are now 100 arcseconds from the center of th

e nebula are moving away from the center by about 0.11 arcsecond per year. Use that information to estimate the year in which the explosion ought to have been observed.
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer:

Estimation: year 1110.

Explanation:

We need to know how much time it takes to move 100 arcseconds if it moves at 0.11 arcsecond per year. Similarly to any velocity equation v=\frac{d}{t}, where in our case the distances are angular, we will obtain the time by doing:

t=\frac{d}{v}=\frac{100arc}{0.11arc/year}=909 years

Which, considering from 2019, the explosion ought to have been observed around 1110 (in reality it was observed by Chinese astronomers in 1054).

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The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he
schepotkina [342]

Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

7 0
3 years ago
Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The
tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q  x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q  /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0
3 years ago
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
A book sits on a bookshelf without moving until a student picks it up. Which law best explains why the book remains at rest unti
svetoff [14.1K]

Answer:

Newtons first law

Explanation:

object in rest stays at rest

object in motion stays in motion

8 0
3 years ago
Explain what differentiates the Earth’s crust and lithosphere.
Allisa [31]

Answer:

What is the difference between the crust and lithosphere? The crust (whether continental or oceanic) is the thin layer of distinctive chemical composition overlying the ultramafic upper mantle. ... The lithosphere is the rigid outer layer of the Earth required by plate tectonic theory.

Explanation:

Hope this helps!

4 0
3 years ago
Read 2 more answers
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