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kari74 [83]
3 years ago
5

An escalator carries you from one level to the next in an airport terminal. The upper level is 4.4m above the lower level, and t

he length of the escalator is 7.3m . Assume that your mass is 63kg .Part AHow much work does the up escalator do on you when you ride it from the lower level to the upper level?Express your answer to two significant figures and include the appropriate units.W =___________.Part BHow much work does the down escalator do on you when you ride it from the upper level to the lower level?Express your answer to two significant figures and include the appropriate units.W =___________.
Physics
1 answer:
horrorfan [7]3 years ago
6 0

Answer:

Part a)

W = 2716.5 J

Part B)

W = -2716.5 J

Explanation:

Part A)

While escalator is moving up work done to move the person upwards is given as

W = mgh

here we know that

m = 63 kg

h = 4.4 m

now we have

W = 63 \times 9.8 \times 4.4

W = 2716.5 J

Part B)

Work done while we move from up to down

So we have

W = - mgh

so we have

W = -2716.5 J

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kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0
3 years ago
On a balanced seesaw, a boy three times as heavy as his partner sits
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Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

W_1 d_1 = W_2 d_2

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

W_1 = 3 W_2

If we substitute this into the equation, we find:

(3 W_2) d_1 = W_2 d_2

and by simplifying:

3 d_1 = d_2\\d_1 = \frac{1}{3}d_2

which means that the boy sits at 1/3 the distance from the fulcrum.

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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
Read 2 more answers
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