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Artyom0805 [142]
3 years ago
7

Can someone plz tell me how to solve this step by step

Physics
1 answer:
avanturin [10]3 years ago
7 0

Answer:

Explanation:

T=2π√l/g

to solve this for l we have to remove the square root in order to finish square root we have to take square on both side so the equation become

T²=4π²l/g

multiplying g on both side

T²g=4π²l

now dividing 4π² on both side

T²g/4π²=l

so l=T²g/4π²

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Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is mo
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considering upward direction to be positive

N-mg=ma

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A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the
lina2011 [118]

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The the analysis for the free fall part should be done under the constant acceleration.

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8 0
2 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
Read 2 more answers
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