A demographic tool used to predict a population's future growth is

Quel est le gaz présent dans la couche d’ozone

stepan [7]

Answer:

L'ozone est produit à partir du dioxygène, composé de deux atomes d'oxygène(O2). Aux altitudes supérieures à 30 km, le rayonnement solaire possède encore une énergie suffisante pour casser une partie des molécules de dioxygène et libérer les atomes

3 0

3 years ago

What is the source of the radioactive nuclei present in spent fuel rods?

Debora [2.8K]

The answer is <span>nuclear fission. T</span>he source of the radioactive nuclei present in spent fuel rods is nuclear fission. Nuclear fission<span> is the </span>process<span> in which a large nucleus splits into two smaller nuclei with the release of energy. </span>

6 0

3 years ago

A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in

Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

$\boxed{u \ or \ v_i = initial \ velocity}$
$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$
$\boxed{a = acceleration \ (constant)}$
$\boxed{d = distance \ travelled}$

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (in m/s²)
u = initial velocity
v = final velocity
d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

$\boxed{ \ 0 = v^2 + 2ad \ }$

$\boxed{ \ v^2 = -2ad \ }$

Both sides are divided by -2d, we get $\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }$

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

$\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }$

Here d' is the stopping distance that we want to look for.

$\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }$

We crossed out 2 in above and below.

$\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }$

We multiply both sides by d.

$\boxed{ \ v^2 d' = 49.0v^2 d \ }$

We crossed out v^2 on both sides.

$\boxed{\boxed{ \ d' = 49.0d \ }}$

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>

Determine the acceleration of the stuffed bear brainly.com/question/6268248
Particle's speed and direction of motion brainly.com/question/2814900
About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.