A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of T = 0.033 s that is increasing at the rate of
1.26×10−5s/y
. (a) What is the pulsar’s angular acceleration α? (b) If α is constant, how many years from now will the pulsar stop rotating? (c) The pulsar originated in a supernova explosion seen in the year 1054. Assuming constant α, find the initial T.

α

b)t = 2619.8 years

c) T(i) =  0.0208536s  

Explanation:

Given that ,

The period of pulsar T = 0.033 s

The period increase at a rate of   = 1.26×10⁻⁵s/y

a) Pulsar angular speed is

ω = θ / T = 2π / T

θ = one revolution of pulsar

T = period of pulsar

Pulsar angular acceleration is given by

b) ω₀ = θ / T

0 = 190.4 - 2.303 × 10⁻⁹

c) 2018 - 1054

= 964years

The pulsar is originated in a supernova explosion 964 years ago.

then the initial period of pulsar

T(i) =  0.0208536s

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

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