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Dovator [93]
3 years ago
13

Determine the resultant moment produced by the load and the weights of the tower crane jibs about point A and about point B. Exp

ress your answers using three significant figures separated by a comma.
Physics
1 answer:
grigory [225]3 years ago
5 0

Answer:

(M)_A = (M)_b = 76027.5 Nm

Explanation:

Step 1:

- We will first mark each weight from left most to right most.

                                 Point:                       Weight:

                                  G_3                          W_g3 = 6 Mg*g

                                  G_2                          W_g2 = 0.5 Mg*g

                                  G_1                           W_g1 = 1.5 Mg*g

                                 Load                           W_L = 2 Mg*g

Step 2:

- Set up a sum of moments about pivot point B, the expression would be as follows:

                   (M)_b = W_g3 * 7.5 + W_g2*4 - W_g1*9.5 - W_L*12.5

Step 3:

- Plug in the values and solve for (M)_b, as follows:

                   (M)_b = 9.81*10^3( 6* 7.5 + 0.5*4 - 1.5*9.5 - 2*12.5)

                   (M)_b = 9.81*10^3 * (7.75)

                   (M)_b =  76027.5 Nm

Step 4:

- Extend each line of action of force downwards, we can see that value of each force does not change also the moment arms of each individual weight also remains same. Hence. we can say that (M)_A = (M)_b = 76027.5 Nm

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AysviL [449]

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

P = \frac{E}{t}

where;

P is power in watts

E is energy in Joules

t is time in seconds

P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W

The average current that this cell phone draws when turned on:

P = IV

I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

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3 years ago
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garik1379 [7]

Answer:

D.

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A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

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3 years ago
In your textbook reading Chapter 26, the author suggests that an electric vehicle (EV) fleet can be used as a kind of distribute
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Answer:

Answer for the question is given in the attachment.

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Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

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\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

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\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

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