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sertanlavr [38]

3 years ago

14

While a gymnast is in the air during a leap, which of the following quantities must remain constant for her?A) Angular momentum

about her center of massB) positionC) velocityD) momentumE) angular velocity

1 answer:

Alinara [238K]3 years ago

7 0

While a gymnast is in the air during a leap, her Angular momentum about her center of mass<em> (A)</em> and her linear momentum<em> (D)</em> must remain constant for her.

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The acceleration of an object is constant when its velocity is :

emmasim [6.3K]

Answer:

B. changing by a constant amount each second

Explanation:

thats my answer

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3 years ago

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If an astronaut has a mass of 80kg on earth, what is their mass on the moon?

stealth61 [152]

Answer:

130 N basically

Explanation:

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3 years ago

I need help with 2 questions . Please help c:

dezoksy [38]

A) According to the nebular theory, the Solar System formed from a huge gaseous nebula which at a certain point was perturbated. Atoms and molecules started colliding, forming planetesimals (a sort of big rocks). The planetesimals were attracted to each other by gravity, forming bigger warm almost spherical objects called protoplanets, which at the end cooled down forming planets.
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3 years ago

earnstyle [38]

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

$x=0.150m-0.115m=0.035m$

And spring constant is $k=51.0N/m$

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3 years ago

4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of

Vesnalui [34]

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle ( $\Phi$ ) = 20°

Pinion speed ( $n_p$ ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear ( $N_G$ ) = 50

Teeth on pinion ( $N_p$ ) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

= $\frac{20}{5}$

= 4 in

From the values of Lewis Form Factor Y for ( $n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

= 2.0472

However, there is need to get the value of the tangential load $(W^t)$ , in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

$W^t = \frac{63025*30}{2000*\frac{4}{2}}$

$W^t = 472.69 lbf$

Finally, the bending stress is calculated via the formula:

$\sigma = \frac{K_vW^tp}{FY}$

$\sigma = \frac{2.0472*472.69*5}{1*0.321}$

$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

4 0

3 years ago

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