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inysia [295]
3 years ago
7

A paleontologist estimates that when a particular rock formed, it contained 12 mg of the radioactive isotope potassium-40, which

has a half-life of 1.26 billion years. The rock now contains 3 mg of the isotope. About how old is the rock? Enter in billions of years.
Physics
1 answer:
leva [86]3 years ago
7 0

Answer:

t = 2.52 billion \:years

Explanation:

As we know by radioactivity law

N = N_o e^{-\lambda t}

so here we will have

N = 3 mg

N_o = 12 mg

now we will have

3 = 12 e^{-\lambda t}

\lambda t = ln 4

now we also know that

\lambda = \frac{ln2}{1.26 \times 10^6 yrs}

t = 1.26\times 10^6\times \frac{ln4}{ln2}

t = 2.52 billion \:years

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Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

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Why are different colors used to represent the floor of the ocean?
ZanzabumX [31]
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5 0
3 years ago
Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl
UkoKoshka [18]

Answer:

a) W_g =61.25J

b) W_k = -46.25J

c) W_N = 0

d) W_g would be the same.

   W_k would decrease.

   W_N would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J

b) The friction force is equal to the normal force times coefficient of friction.

F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

h = x\sin(\theta)

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of x\sin(\theta).

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0
3 years ago
an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying a
mart [117]

Answer:

331.7m/s

Explanation:

Given parameters:

Initial velocity  = 100m/s

Acceleration  = 50m/s²

Distance  = 1km   = 1000m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we have to apply the right motion equation shown below;

     v²  = u²  + 2aS

 v is the final velocity

 u is the initial velocity

 a is the acceleration

 S is the distance

 Now insert the parameters and solve;

     v² = 100² + (2 x 50 x 1000)

     v² = 110000

     v = √110000  = 331.7m/s

4 0
3 years ago
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