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BigorU [14]
3 years ago
8

Choose a sport you like, such as football or track Describe how a high level of flexibility would affect your performance in tha

t sport. Consider the benefits in many aspects of the sport. PLZZZ HELP ME OUT
Physics
2 answers:
irinina [24]3 years ago
6 0

Answer:

Swimming

Explanation:

An increase of flexibility for swimming helps elongate the swimmer to propel faster and glide through the water easier. If the swimmer is inflexible, they won't be able to stretch as far as they should. But, if they were flexible, it'd be easier and much simpler.

Brainlist pls!

Sveta_85 [38]3 years ago
5 0

Answer:

I like MMA, high level of flexibility would affect the performance in a positive way. MMA requires you to be flexible for both offense and defense, and have a wider range of techniques than the person who is less flexible, therefore having more advantage. The benefits of MMA as a sport is you learn to fight, you get fit, and you are more flexible.

Explanation:

Since it can be any sport, I did MMA, and got 100% I hope this is useful to you or someone else :)

You might be interested in
Which of these rotational quantities is analogous to mass in a linear system?
Rom4ik [11]

Answer:

d

Explanation:

i think it is rotational inertia

because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.  

hope it's right & helps !!!!!!!!!

5 0
2 years ago
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject
kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

Radius_{mean} = 60\ mm

Diameter_{mean} = 120\ mm

Load = 200 N

We need to calculate the deflection

Using formula of deflection

\delta=\dfrac{8pD^3n}{Cd^4}

Put the value into the formula

\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}

\delta =34.56\ mm

Hence, The deflection of the spring is 34.56 mm.

4 0
3 years ago
A student uses 60 Newtons of force to climb 18 meters in 6 seconds. Calculate her Power.
puteri [66]

Answer:

15

Explanation:

P=W/T

T=6sec

W=?

F=60N

S=18m

W=F X S. .s indicate displacement

W=60x18

W=108

So p=108 j/6sec

P=15watt

5 0
2 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the
Luda [366]
  • Answer:

<em>12,25 km/h</em>

<em>≈ 3,4 m/s </em>

  • Explanation:

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>= 12,25km/h</em>

<em>or</em>

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>1h = 60m×60s = 3600s</em>

<em>= 12250m/3600s</em>

<em>≈ 3,4 m/s </em>

5 0
3 years ago
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