Can someone explain to how to calculate this

Write a hypothesis about how the number of half-lives affects the number of radioactive atoms. Use the "if . . . then . . . beca

Roman55 [17]

Answer:

Explanation:

Answer:

Explanation:

The half life is the time taken for half of a radioactive substance to disintegrate.

The shorter the half life, the larger the decay constant and the faster the decay process.

For a very large half life, it would take a very long time for the radioactive nuclide to decay to half.

With each half life reached, a new set of daughter cell is formed. Atoms that have short half life would decay rapidly. Every radionuclide has its own characteristic half-life.

If the number of half-lives increases, then the number of radioactive atoms decreases, because approximately half of the atoms' nuclei decay with each half-life. With this observation, we can hypothesise and conduct experiment to support the assertion that as the number of half-lives increases then the number of radioactive atoms decreases.

8 0

3 years ago

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The position of a certain airplane during takeoff is given by x=1/2 *bt2, where b = 2.0 m/s2 and t = 0 corresponds to the instan

Serhud [2]

Answer:

1362000 kgm/s

Explanation:

So the total mass combination of the plane and the people inside it is

M = 35000 + 160*65 = 45400 kg

After 15 seconds at an acceleration of 2 m/s2, the plane speed would be

V = 2*15 = 30 m/s

So the magnitude of the plane 15s after brakes are released is

MV = 45400 * 30 = 1362000 kgm/s

5 0

3 years ago

. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of

Rina8888 [55]

The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

$F_f = \mu W$

where

$\mu$ is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

$\mu = 0.55$ (coefficient of friction)

$W=580 N$ (weight of the leaves)

Substituting, we find:

$F=(0.55)(580)=319 N$

Learn more about force of friction:

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#LearnwithBrainly

7 0

3 years ago

A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia

Bingel [31]

Answer

given,

diameter of the pipe is = (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

$F = \dfrac{mv^2}{r}$

equating both the force equation

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{gr}$

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s² or g = 9.8 m/s²

$v = \sqrt{32 \times 7}$

v = 14.96 ft/s

or

$v = \sqrt{9.8 \times 2.135}$

v = 4.57 m/s

5 0

3 years ago

Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr > Vw.

miss Akunina [59]

Answer:

Δt = $\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}$

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 + Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

$\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}$ = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

8 0

3 years ago