1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
pashok25 [27]
2 years ago
10

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

Physics
1 answer:
soldi70 [24.7K]2 years ago
8 0

b i think brainliest pls

You might be interested in
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
Other questions:
  • Which scale of measurement measures the magnitude or strength of an earthquake based on seismic waves?
    8·2 answers
  • Suppose you have 150. ml of a substance with density of 42.0 g/dm3. convert the volume from milliliters to liters to cubic decim
    9·1 answer
  • If "38 %" of the light passes through this combination of filters, what is the angle between the transmission axes of the filter
    5·1 answer
  • A string is wound tightly around a fixed pulley having a radius of 5.0 cm. as the string is pulled, the pulley rotates without a
    12·2 answers
  • What are the public policy alternatives to hunger
    11·1 answer
  • Examples of applied force
    12·2 answers
  • Which of the following is not included in impact damage?
    7·2 answers
  • A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav
    13·1 answer
  • What describes the main benefit to organisms that reproduce sexually?
    10·2 answers
  • Which kinetic chain checkpoint should be observed carefully because it controls the movement of the lower extremities? The hip T
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!