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Zepler [3.9K]
3 years ago
6

HELP PLEASECan someone please explain to me why the answer would be 2.45m/s?

Physics
2 answers:
trapecia [35]3 years ago
8 0

Solution with explanation is given below in attachment.

marissa [1.9K]3 years ago
4 0
It's basically asking for the acceleration. When an object is in free fall, the acceleration is just gravity. so in this case ∆v = 2.45 because it is the gravity on that planet.
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It is symbols to make it easier and to save space! 
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H. The length of the shadow is different in evening and in the day. Justify​
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the shadows are exactly the same length in the morning as they are in the evening.

 is so obvious it’s that when the sun is low you get long shadows and when the sun is up in the sky like in the noon the shadow is shorter.

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An airplane with a mass of 5,000 kg needs to accelerate 5 m/s2 to take off before it reaches the end of the runway. How much for
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Answer:

<h2>25000 N</h2>

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force = mass × acceleration

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<h3>25000 N</h3>

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Which of the following is NOT an example of a correct interaction between the
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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
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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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