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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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Which state of matter has atoms that are spread out and bouncy?

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The stage where atoms are spread out and bouncy is the gas stage.

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2 years ago

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

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3 years ago

A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.

Zolol [24]

These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is $45^{\circ}$ .

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
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Finally, the efficiency $\epsilon$ of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

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3 years ago

a car traveling north at 17.7 m/s. after 12 seconds its velocity is 14.1 m/s in the same direction. find the magnitude and direc

elixir [45]

Acceleration is equal to the change in velocity per unit time. In this case, velocity has changed from 17.7 m/s to 14.1 m/s, which is a decrease of 3.6 m/s. This takes place in 12 seconds, so the average acceleration is 3.6 m/s / 12s = 0.3 m/s^2. The direction of this acceleration is south, since the car's northbound speed is decreasing.

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3 years ago

Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K

Heat capacity at constant pressure = 6.9 cal/mol.K

Initial temperature = 300 K