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3 years ago

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A wheel has an initial clockwise angular velocity of 10 rad/sec and a constant angular acceleration of 3 rad/sec2. What time is

required to acquire a clockwise angular velocity of 15 rad/sec? a)- 1.67 sec b)- 8.2 sec c)- 12 sec d)- 3 sec

1 answer:

tino4ka555 [31]3 years ago

7 0

0 duuuuuuuuuuuuuuuuuuuuh

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B. The space between the galaxy is getting better

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In the figure below, block A weighs 20 lb , while block B weighs 10 lb . Friction between the surfaces of the two blocks may be

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Answer:

As P is continually increased, the block will now slip, with the friction force acting on the block being: f = muK*N, where muK is the coefficient of kinetic friction, with f remaining constant thereafter as P is increased.

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2 years ago

ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in

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Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

$d_{p} =mN_{p}$

Where

m = module = 5

Np = number of teeth of pinion = 26

$d_{p} =5*26=130mm$ = 0.13 m

The pitch line velocity is equal to:

$V_{t} =\frac{d_{p}*2*\pi *w_{p} }{120}$

Where

wp = speed of the pinion = 1800 rpm

$V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s$

The factor B is equal to:

$B=\frac{(12-Q_{v})^{2/3} }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396$

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

$K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832$

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

$P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}$

The bending stress is equal to:

$\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}} \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa$

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3 years ago

Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools

dusya [7]

Answer:

Both Technician A and Technician B are correct

Explanation:

Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.

Technician B is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.

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3 years ago

A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr

seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

$\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625$

Weight transfer is given as follows;

$W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N$

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

$F_R$ × 1.3 = 1.7 × $F_F$

$F_R$ + $F_F$ = 18000

$F_R + \dfrac{1.3 }{1.7} \times F_R = 18000$

$F_R$ = 18000*17/30 = 10200 N

$F_F$ = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel $F_R$ = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel $F_F$ = 7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

$\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625$

The least coefficient of friction, μ = 0.625.

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4 years ago