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3 years ago

What is the half-life of this radioactive isotope that decreases to one-fourth its original amount in 18 months?

1 answer:

4 0

The half-life of this radioactive isotope is 9 months for it decreases to one-fourth its original amount in 18 months. Half-life means a certain length of time after which half of the amount of radioactive element has decayed.
Therefore, the half-life of this particular isotope is 18 months. After 18 months the isotope was at one-half, then another 18 months, the isotope was at one-fourth, which is 9 months where one-fourth is one-half of one half.

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The force exerted on the tires of a car that directly accelerate it along a road is exerted by the

azamat

The force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

<h3>What is force?</h3>

Force is defined as the product of mass and acceleration of an object.

Friction is defined as the force that resists the movement of an object over another.

Therefore, the force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

Learn more about force here:

brainly.com/question/12970081

#SPJ12

7 0

2 years ago

A car traveling at 23m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is the acceleration?

Lena [83]

The car has an initial velocity $v_0$ of 23 m/s and a final velocity $v$ of 0 m/s. Recall that for constant acceleration,

$v=v_0+at$

The car stops in 7 s, so the acceleration is

$0\,\dfrac{\mathrm m}{\mathrm s}=23\,\dfrac{\mathrm m}{\mathrm s}+a(7\,\mathrm s)$

$\implies a\approx-3.29\,\dfrac{\mathrm m}{\mathrm s^2}$

7 0

3 years ago

Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the

masya89 [10]

The crate’s acceleration is gravity= 9.81m/s^2
The ball’s acceleration is also gravity=9.81m/s^2

8 0

3 years ago

A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in

Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J

But ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0

3 years ago

Read 2 more answers

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

3 years ago