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tatiyna
3 years ago
13

What is the difference between a vigorous sport and vigorous recreation ?

Physics
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer:

Explanation:El ejercicio vigoroso previene en mayor medida el síndrome metabólico (un conjunto de enfermedades que aumentan el riesgo cardiovascular         )

mientras que una reacción vigorosa se   produce entre el aluminio y el gas cloro. Como consecuencia de la gran cantidad de energía liberada se producen luz y calor

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An object that is accelerating may be
pickupchik [31]
The answer of this question is D. All of the above

Acceleration happen when an obeject change its velocity. It has nothing to do with speed.

The huge misconception about acceleration is when we thought it only aply if we increase our speed ( in a sport match, sportcaster often describe acceleration as an increase in players speed)

slower, faster, right , left, it does not matter, as long as that object change its velocity, it accelerates
8 0
2 years ago
Which is a unit to measure air pressure?
nata0808 [166]

Answer:

c) cubic centimetre is it's answer..

8 0
2 years ago
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A nation undergoing the transition from undeveloped to industrialized is commonly dependent on this to supply its energy needs.
algol [13]
A) obviously :)))))))))
7 0
2 years ago
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When NASA was communicating with astronauts on the Moon, the time from sending on the Earth to receiving on the moon was 1.33 s.
frez [133]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description, which determine the velocity, such as the displacement of a particle as a function of time, that is to say

v = \frac{x}{t}\rightarrow x = v*t

Where,

x = Displacement

v = Velocity

t = Time

Our values are given as,

v=3*10^8m/s

t = 1.33 s

Replacing we have that,

x=v*t

x=(3*10^8)(1.33)

x = 399'000.000m

Therefore the distance from Earth to the Moon is 399.000 km

5 0
3 years ago
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

4 0
3 years ago
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