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garik1379 [7]
3 years ago
13

A radioactive substance decays according to the formula w = 20e¡kt grams where t is the time in hours. a find k given that after

50 hours the weight is 10 grams. b find the weight of radioactive substance present at: i t = 0 hours ii t = 24 hours iii t = 1 week. c how long will it take for the weight to reach 1 gram? d find the rate of radioactive decay at: i t = 100 hours ii t = 1000 hours. e show that dw dt is proportional to the weight of substance remaining.
Physics
1 answer:
blagie [28]3 years ago
5 0
W=20 e(-kt)  
A. Rearranging gives k= -(ln(w/20)/t 
 Substituting w= 10 and solving gives k=0.014 
 B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g 
 C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours. 
 D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h 
 E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW
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False

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9 = kq1q2/1²

9 = kq1q2 ..... 1

If the charges are pushed to a 3 meter separation, then;

F =  kq1q2/3²

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Divide both equations;

9/F = (kq1q2)/ kq1q2/9

9/F =  kq1q2 * 9/ kq1q2

9/F = 9

F = 9/9

F = 1N

Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False

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When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
3 years ago
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