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3 years ago

When does the radioactive decay of a radioisotope stop? Give one example.

Physics

2 answers:

Romashka-Z-Leto [24]3 years ago

5 0

Answer:

When does the radioactive decay of a radioisotope stop? Give one example. An unstable isotope continues the decay process until it reaches a stable form. One example is the decay of carbon-14 to nitrogen-14.

Explanation:

Rudik [331]3 years ago

4 0

Answer:

Sample response: An unstable isotope continues the decay process until it reaches a stable form. One example is the decay of carbon-14 to nitrogen-14.

Explanation:

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A perpetual-motion machine can never be built because it is not possible to eliminate...

bazaltina [42]

The answer is: [C]: "elasticity" .
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4 0

3 years ago

Read 2 more answers

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

F = m a

a = F / m

We use kinematics to find lips

v = v₀ + a t

v = v₀ + (F / m) t

The moment is defined by

p = m v

The moment change

Δp = m v - m v₀

Let's replace the speeds in this equation

Δp = m (v₀ + F / m t) - m v₀

Δp = m v₀ + F t - m v₀

Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

8 0

3 years ago

A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.

scoundrel [369]

Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB