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Alexandra [31]
4 years ago
7

Why should flexibility exercises be done in conjunction with strength-building exercises?

Physics
2 answers:
bixtya [17]4 years ago
7 0
<span>Flexibility exercises should be done in conjunction with strength-building exercises because: 

</span><span>They work together to maintain your overall level of fitness.
</span>
I hope my answer has come to your help. God bless and have a nice day ahead!
agasfer [191]4 years ago
7 0

Answer:

The answer is flexibility exercise allows muscles to perform strength exercises with greater capacity.

Explanation:

The strength-building exercise should be done in conjunction with the flexibility exercises because it allows us to protect our muscles and joints from possible injuries, in addition to providing us with a greater and better range of movement since a relaxed muscle has more ease of performing a rapid contraction and therefore more possibility of developing a greater force.

There is a direct relationship between flexibility and the ability to execute movements with power since a flexible muscle has an adequate capacity to exert its full power.

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jek_recluse [69]

Answer:

D

Explanation:

D it generates pressure by fusing hydrogen.

4 0
3 years ago
James has a mass of 98 kg and Basma has a mass of 59 kg. James is running at 3.0 m/s, while Basma is running at 4.0 m/s.
masha68 [24]

(a) James has the most momentum which is 294 kgm/s.

(b) The resultant force acting on Basma is 90.78 N.

(c) The time taken for James to stop is 3.2 seconds.

<h3>Momentum of each person</h3>

Momentum of James: P = mv = 98 x 3 = 294 kgm/s

Momentum of Basma: P = mv = 59 x 4 = 236 kgm/s

<h3>Resultant force of Basma</h3>

F = ma = mv/t = P/t = 236/2.6 = 90.78 N

<h3>Time for James to stop</h3>

F = P/t

t = P/F

t = 294/90.78

t = 3.2 s

Learn more about momentum here: brainly.com/question/7538238

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4 0
2 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
What type of energy conversion occurs at the moment fireworks explode?
Paladinen [302]

chemical energy converts  into light, heat and sound  energy

6 0
4 years ago
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There are four springs stretched by the same mass.
kirill115 [55]

Answer:

That would be Spring C

Explanation:

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