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alexandr402 [8]
3 years ago
9

Elabora una tabla, como la del ejemplo, con los resultados obtenidos en los test que desarrollaste en actividades anteriores. Lu

ego, responde: ¿qué indican esos resultados con respecto a tu condición física y cómo se relacionan con tu salud? A partir de tus respuestas, plantéate el reto de mejorar tu condición física manteniendo o mejorando tu rutina de ejercicios o la práctica permanente de actividad física.
Physics
1 answer:
rodikova [14]3 years ago
8 0

Explanation:

Clear rendering reads;

"Make a table, like the one in the example, with the results obtained in the tests you carried out in previous activities. Then answer: What do these results indicate regarding your physical condition and how do they relate to your health? From your answers, consider the challenge of improving your physical condition by maintaining or improving your exercise routine or permanent practice of physical activity".

So the incomplete text above it seems you've been instructed to perform an experiment and then state your result/analysis.

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A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0
3 years ago
Three currents are passing through a surface bounded by a closed path. The currents have different values and directions. Accord
GaryK [48]

Answer:

See Explanation Below

Explanation:

This question is incomplete.

I'll answer this question on general terms. You'll get your result if you apply the steps I'll highlight below.

To start with; what's Ampere law;

It states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.

The simple translation of this law is that, the sum of current in the close loop gives us the desired result.

Rephrasing your question;

Three currents, (I1 = +3A, I2 = +4A and I3 = -5A) are passing through a surface bounded by a closed path. The currents have different values and directions. According to Ampere’s law, what is the value of I on the right side of this equation?

First, we take note of the signs in front of the given currents.

The negative sign in front of I3 means that; it is moving in opposite direction of I1 and I2.

To calculate the value of I.

The value of I is the sum of the three currents:

i.e. 3A + 4A - 5A

I = 2 A

7 0
3 years ago
PLEASE HURRY THIS IS URGENT
Irina-Kira [14]

3. Rubbing Sticks together to create a fire.


5. Gravity.

7 0
3 years ago
What are the atmospheric conditions over Minneapolis,Minnesota? Check all that apply
Masja [62]

Answer:

the answer is  A & C & E just did the lab

Explanation:

3 0
3 years ago
Read 2 more answers
A fluid, with a density of rho = 1165 kg/m3, flows in a horizontal pipe. In one segment of the pipe the flow speed is v1 = 4.53
Elza [17]

Answer:

The pressure difference between two pipe is 1.01 \times 10^{4} Pa

Explanation:

Density \rho = 1165 \frac{kg}{m^{3} }

Speed in one pipe v_{1} = 4.53 \frac{m}{s}

Speed in second pipe v_{2} = 1.77 \frac{m}{s}

According to the bernoulli equation,

The pressure difference is given by,

     P = \frac{1}{2} \rho v^{2}

P_{2} - P_{1} = \frac{1}{2}  \rho (v_{1}^{2}  - v_{2}^{2}  )

P_{2} - P_{1} = \frac{1}{2} \times 1165 \times[ (4.53)^{2}- (1.77)^{2}]

P_{2} - P_{1} = 10128.51

P_{2} - P_{1} = 1.01 \times 10^{4} Pa

Therefore, the pressure difference between two pipe is 1.01 \times 10^{4} Pa

6 0
3 years ago
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