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kvasek [131]
3 years ago
13

A sample contains 20 kg of radioactive material. The decay constant of the material is 0.179 per second. If the amount of time t

hat has passed
is 300 seconds, how much of the of the original material is still radioactive? Show all work
Physics
1 answer:
Tju [1.3M]3 years ago
6 0

Answer:

There are 9.537\times 10^{-23} kilograms of radioactive material after 300 seconds.

Explanation:

From Physics we know that radioactive materials decay at exponential rate, whose differential equation is:

\frac{dm}{dt} = -\lambda\cdot m (1)

Where:

\frac{dm}{dt} - Rate of change of the mass of the radioactive material, measured in kilograms per second.

m - Current mass of the radioactive material, measured in kilograms.

\lambda - Decay constant, measured in \frac{1}{s}.

The solution of the differential equation is:

m(t) = m_{o}\cdot e^{-\lambda\cdot t} (2)

Where:

m_{o} - Initial mass of the radioactive material, measured in kilograms.

t - Time, measured in seconds.

If we know that m_{o} = 20\,kg, \lambda = 0.179\,\frac{1}{s} and t = 300\,s, then the initial mass of the radioactive material is:

m(t) = (20\,kg)\cdot e^{-\left(0.179\,\frac{1}{s} \right)\cdot (300\,s)}

m(t) \approx 9.537\times 10^{-23}\,kg

There are 9.537\times 10^{-23} kilograms of radioactive material after 300 seconds.

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A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable
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Answer:

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Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

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Solution :

Magnetic field at the centre due to $I_P$ :

$B_1 = \frac{\mu_0 I_P}{2 \pi d}$

$B_1 = \frac{4 \pi \times 10^{-7} \times 0.2}{2 \pi d}$

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Its direction will be downwards in the plane of the paper.

Magnetic field at the centre due to $I_Q$ :

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Now the resultant of the magnetic field at the centre.

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