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Elena-2011 [213]
3 years ago
7

Qué sistemas del organismo afecta la falta de ejercicio?

Physics
1 answer:
Alex_Xolod [135]3 years ago
4 0

Answer:

ndmnk.nhlwhfliehfhewfiberjihfbjewbfujebwgwuerkjfhwg.augfukGARWufg.hbjkrwfhogujbr

Explanation: fhulhfuwgeuikBWGFEIUgwfk.ewgukhGFHGUE<FR<HKEl8oiufkhy48e6r3yuogfkh

You might be interested in
As the velocity of the sound source approaches the speed of sound, the wave fronts ahead of the object begin to appear as:
sineoko [7]

Answer:

The correct answer is D

shock waves

Explanation:

When a body approaches the speed of sound in the media, the waves it emits are getting closer and closer to each other, the envelope (sum) results in a kind of cylindrical cone that is similar to that formed in a supersonic process

The correct answer is D

When the speed of sound is exceeded, the body goes faster than the wave, so the so-called shock waves are formed.

  They carry a lot of energy

4 0
3 years ago
If a girl walks 6km due west changes directions and walk another 5km due north. Find her displacement in both magnitude and dire
xenn [34]

Answer:

the girl's displacement in both magnitude and direction is 7.81 m at 50.2⁰ North west.

Explanation:

Given;

6km due west, and

5km due north

The magnitude of her displacement is calculated by forming a right angled triangle. The hypotenuse side of the triangle is the girl's displacement.

d² = 5² + 6²

d² = 25 + 36

d² = 61

d = √61

d = 7.81 m

The direction of the girl is calculated as;

tan \ \theta = \frac{6}{5} \\\\tan \ \theta = 1.2\\\\\theta = tan^{-1} (1.2)\\\\\theta = 50.2^0

Therefore, the girl's displacement in both magnitude and direction is 7.81 m at 50.2⁰ North west.

4 0
3 years ago
If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will
astraxan [27]

Answer:

Therefore the answer is the precision in the speed DECREASES

Explanation:

In quantum mechanics, we have the uncertainty principle that establishes that when the accuracy of the position increases the accuracy  the speed decreases, being related by the expression

               Δx Δv ≥  h'/ 2

                h' = h/2π

Therefore the answer is the precision in the speed DECREASES

7 0
4 years ago
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
3 years ago
A car is moving north on a freeway. If a bug is flying south on the freeway, is the momentum of the bug positive or negative?
faltersainse [42]

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

3 0
2 years ago
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