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iogann1982 [59]
3 years ago
15

The coach has not seen improvement as expected in his team and plans to change practice session. To help the players he changes

the Frequency, Intensity, Time and Type of their exercise and uses the fitness principle of frequency
overload
progression
reversibility​
Physics
1 answer:
Allushta [10]3 years ago
7 0

Answer:

progession

Explanation:

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7. An astronaut weighs 900N on earth. On the moon, he weighs 150N. Calculate the
noname [10]

Answer:

Moons’ gravitational strength = weight of astronaut on the moon / mass of astronaut.

= 150 / 90 = 1.67 Nkg-1

Explanation:

7 0
3 years ago
The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro
Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

Pressure ,P = 1 atm

As we know that  psychrometric chart are drawn at constant pressure.

From the diagram

ω= specific humidity

Lets take these two lines   Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P

From chart at point P

a)

Specific humidity,ω = 0.00922 kg/kg

b)

The enthalpy ( h)

h=47.59 KJ/kg

c)

The relative humidity, RH

RH= 49.58 %

d)

Specific volume ,

v= 0.853 m³/kg

5 0
3 years ago
Which event is part of a convince part of a current in the air
Effectus [21]
The answer is B for sure !
4 0
3 years ago
Read 2 more answers
Consider a block on frictionless ice. Starting from rest, the block travels a distance din
sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{F}{m}

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

\displaystyle d = vo.t+\frac{at^2}{2}

If the block starts from rest, vo=0:

\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

\displaystyle d = \frac{\frac{F}{m}t^2}{2}

Simplifying:

\displaystyle d = \frac{Ft^2}{2m}

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

\displaystyle a'=\frac{4F}{m}

And the distance is now:

\displaystyle d' = \frac{4Ft^2}{2m}

Dividing d'/d:

\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}

Simplifying:

\displaystyle \frac{d' }{d}=4

Thus:

d' = 4d

The distance is now 4d

3 0
2 years ago
*15 points*
Svetradugi [14.3K]
Same as the other person most likely would be 20 times louder as your answer
5 0
2 years ago
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