Se deja caer una caja de madera de 4.5kg de masa desde una altura de 2.25metros .

What is the relationship between temperature and altitude in the stratosphere? (2 points).

stealth61 [152]

As altitude increases, temperature increases. The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>

8 0

3 years ago

Which statements best describe the second stage of cellular respiration? Check all that apply.

grandymaker [24]

Answer:

oxygen combines with small molecules. energy is released and it happens in the cytoplasm

Explanation:

7 0

3 years ago

Read 2 more answers

A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i

Stolb23 [73]

The speed $V_{i}$ of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.

We are given that-

the mass of the elevator (m) = 1000 kg ;

the distance the elevator decelerated to be y = 8m ;

the tension is T = 11000 N;

let us determine the acceleration 'a' by using Newton's second law of motion.

∑Fy = ma

W - T = ma

(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a

9800 - 11000 = 1000

a = - 1.2 m/s²

Using the equation of kinematics to determine the initial velocity.

$V_{f}$ ² = $V_{i}$ ² + 2ay

$V_{i}$ = √ ( 2 x 1.2m/s² x 8 m )

$V_{i}$ = √19.2 m²/s²

$V_{i}$ = 4.38 m/s ≈ 4 m/s

Hence, the initial velocity of the elevator is 4m/s.

Read more about the Equation of kinematics:

brainly.com/question/12351668

#SPJ4

8 0

1 year ago

The equation P^xV^yT^z= constant is Boyle law for what is the values of x,y,z

chubhunter [2.5K]

I learned the equation as P•V=k•T .

So x=1, y=1, and z= -1 .

3 0

2 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago