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3 years ago

The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.

Physics

1 answer:

Rainbow [258]3 years ago

8 0

Answer: $0.0353\ s^{-1}$

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in $2.59\times 10\ s$

Sample at any time is given by

$N=N_oe^{-\lambda t}$

where, $\lambda=\text{decay constant}$

Put values

$\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10$

Taking natural logarithm both side

$\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}$

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Read 2 more answers

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

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Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

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$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

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b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :

Then:

$\sum M _B = 0$

$P*h = mg*0.24$

$h =\frac{45*9.8*0.24}{132.3}$

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

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