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3 years ago

Which is not a component of skill-related fitness?

2 answers:

5 0

You did not give options to your question, but you can use this explanation, it will point you in the right direction.
There are six components of skill related fitness, they are AGILITY, BALANCE, COORDINATION, SPEED, POWER AND REACTION TIME.
These components are not just for sporting activities only, we use them too in our daily activities. Exercise and activities promote these components.

steposvetlana [31]3 years ago

4 0

The answer is heart rate

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Tar is hard to pour because it doesn’t flow easily

IgorC [24]

tar is very thick and viscous making it difficult to pour when cold which is why it's heated to extremely high temperatures when blacktop is being poured

6 0

3 years ago

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Give your answer to 2 dp

AVprozaik [17]

Answer:

67.5

Explanation:

The plane accelerates at 2.7m/s,^2

Time is 25 seconds

The velocity can be calculated as follows

= 25×2.7

= 67.5

Hence the speed f the plane is 67.5

6 0

3 years ago

Consider two planets of mass m and 2m,

Rzqust [24]

Answer:

Part a)

$\frac{F_1}{F_2} = 10.125$

Part b)

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

$\frac{T_1}{T_2} = 9.54$

Explanation:

Part a)

As we know that the gravitational force is given as

$F = \frac{GMm}{r^2}$

so we will have to find the ratio of force on two planets due to star

so here we have

$\frac{F_1}{F_2} = \frac{m_1r_2^2}{m_2r_1^2}$

$\frac{F_1}{F_2} = \frac{m (4.5r)^2}{(2m) r}$

$\frac{F_1}{F_2} = 10.125$

Part b)

Orbital speed is given as

$v = \sqrt{\frac{GM}{r}}$

so the ratio of two orbital speed is given as

$\frac{v_1}{v_2} = \frac{r_2}{r_1}$

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

Time period is given as

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

so the ratio of two time period is given as

$\frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}}$

$\frac{T_1}{T_2} = \sqrt{\frac{4.5r^3}{r^3}}$

$\frac{T_1}{T_2} = 9.54$

8 0

3 years ago

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Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m

Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

$A=\frac{1-V_{1}^{2} }{E_{1} }-\frac{1-V_{2}^{2} }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}$

a) The maximum contact pressure is:

$P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} }) }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa$

The width of contact is:

$b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} }) } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm$

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

6 0

3 years ago

When nuclear fission occurs, some mass is lost. Where does the mass go?

Hoochie [10]

The situation when nuclear fission occurs, some mass is lost is also called the mass defect and as a result the matter turns into energy. If you want to know where the mass does go Albert Einstein will be your helping hand : E = mc^2. Using this formulae you will solve the task and find out the direction of mass.

3 0

3 years ago

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