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3 years ago

15

488 J of work is done to a box which is moved across the floor for a distance of 8.9 m. What net force is required to act on the

box to do this amount of work?

1 answer:

kvv77 [185]3 years ago

8 0

According to the formula
$a = f \times d$
Where a is work, f is force and d is the distance that box was moved over. And from that formula, you can get that f = a/d and that is 54.83N of force

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Question:

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4) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

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Explanation:

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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

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3 years ago

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