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2 years ago

7

a An object is tarown up with a velocity v = 6.02 +7.0j. Calculate the (1) time taken reach the maximum height (ii) the horizont

al range (s = 10m/s2).

2 answers:

AfilCa [17]2 years ago

8 0

Answer:

(i) 0.6s (ii) 8.42m

Explanation:

U² = 6.02² + 7²

U = 9.23

angle of projection

tanø = 6.02/7

ø = 40.7

Time of fligt

t = Usinø/g

t = 9.23 sin 40.7/10

t = 0.6

H range = U²sin2ø/g

H = 9.23²sin 81.4/10

H = 8.42m

Shkiper50 [21]2 years ago

4 0

Answer:

the update is coming down 55th 55 cm long

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Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

$R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N$ ....(I)

Where, N = number of radio active atoms

$t_{\frac{1}{2}}$ =half life

We need to calculate the number of radio active atoms

For $N_{12_{c}}$

$N_{12_{c}}=\dfrac{N_{A}}{M}$

Where, $N_{A}$ =Avogadro number

$N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}$

$N_{12_{c}}=5.02\times10^{22}\ nuclie/g$

For $N_{c_{14}}$

$N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}$

$N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}$

$N_{c_{14}}=6.526\times10^{10}\ nuclei/g$

Put the value in the equation (I)

$R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}$

$R=15.0650\ decay/min g$

100 g carbon will decay with rate

$R=100\times15.0650=1507\ decay/min$

We need to calculate the total half lives

$(\dfrac{1}{2})^{n}=\dfrac{390}{1507}$

$2^n=\dfrac{1507}{390}$

$2^n=3.86$

$n ln 2=ln 3.86$

$n=\dfrac{ln 3.86}{ln 2}$

$n =1.948$

We need to calculate the age of living tree

Using formula of age

$t=n\times t_{\frac{1}{2}}$

$t=1.948\times5700$

$t=11103.6 =11104\ years$

Hence, The age of living tree is 11104 years.

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