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3 years ago

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Which of the following is not a component of fitness? A. Cardiovascular fitness B. Muscular endurance C. Pulmonary endurance D.

Body composition

1 answer:

Ne4ueva [31]3 years ago

7 0

Answer:

C. Pulmonary endurance

Explanation:

I'm pretty sure it's "C" because cardiovascular and pulmonary endurance are the same thing and usually you'd hear cardiovascular more than pulmonary.

Sorry if I'm wrong!

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At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l

Ksju [112]

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

5 0

3 years ago

How do you calculate the net force when there are multiple forces in different directions?

Artyom0805 [142]

To find $F_{net}$ we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force $F_{net} = \sqrt{(3)^2+(4)^2} \\$ which gives $F_{net} = 5$ . We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using $\theta = \tan^{-1} (\frac{4}{3})$ where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.

4 0

3 years ago

According to the law of refraction, light passing from air into a piece of glass at an angle of 30 degrees will cause the light

pantera1 [17]

Answer:

bend toward the normal line

Explanation:

When light passes from a less dense to a more dense substance, (for example passing from air into water), the light is refracted (or bent) towards the normal. In your question the light is moving from rarer to denser medium

4 0

3 years ago

The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo

a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions

4 0

3 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

2 years ago