How does quantum mechanics relate to the universe?

How does the matter and (solar) energy move through the sphere?

Tema [17]

When the Sun's energy moves through space, it reaches Earth's atmosphere and finally the surface. This radiant solar energy warms the atmosphere and becomes heat energy. This heat energy is transferred throughout the planet's systems in three ways: by radiation, conduction, and convection.

5 0

2 years ago

Erica (38 kg ) and danny (43 kg ) are bouncing on a trampoline. just as erica reaches the high point of her bounce, danny is mov

AfilCa [17]

<span>2.5 m/s going upward.
In the situation described, Erica and Danny undergo a non-elastic collision which will conserve their combined momentum. Since Erica is stationary, her momentum is 0. And since Danny is moving upward at 4.7 m/s his momentum is 43 kg * 4.7 m/s = 202.1 kg*m/s. Assuming that both Erica and Danny will be moving as a joined system, their combined mass is 38 kg + 43 kg = 81 kg. Since the momentum will be the same, their velocity will be 202.1 kg*m/s / 81 kg = 2.495061728 m/s. Since we only have 2 significant figures in the provided data, rounding the result to 2 significant figures gives a velocity of 2.5 m/s going upward.</span>

4 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

Tudy the images about geologic time.

Inessa [10]

Answer:

The first flowering plants appeared in the Mesozoic era, not the Paleozoic era.

Explanation:

The Mesozoic era was an era where numerous organisms started to develop in very unique and more advanced ways, both the animals and the plants. In the last period of the Mesozoic, the Cretaceous, the first flowering plants started to appear on the scene. This was revolutionary trait of the plants, and soon these plants started to occupy more and more space and became one of the dominant organisms on the planet. Other important evolution that took place in this period are the appearance of the dinosaurs and the mammals, both becoming the dominant animals on the planet, first the dinosaurs, after that the mammals.

8 0

2 years ago

Read 2 more answers

is using electrolysis to separate water molecules in hydrogen and oxygen atoms , a physical or chemical change?

DedPeter [7]

Technically it's a chemical change even though the process is reversible.

8 0

3 years ago