All categories

3 years ago

A body of weight 300N is lying rough

Engineering

1 answer:

kumpel [21]3 years ago

6 0

Answer:

Horizontal force = 89.2 N

Explanation:

The frictional force = coefficient of friction * magnitude of the force (weight of the body) * cos theta

Substituting the given values, we get -

Frictional Force = 0.3*300 * cos 25 = 89.2 N

Horizontal force = 89.2 N

You might be interested in

Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D

Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ F$

Change in temperature, $\Delta T = T_2-T_1$

$\Delta T = -2.43-87.745$

$\Delta T=-90.175^\circ F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

$=11.401+0.32198(82.898)$

$=38.09297 \ Btu/lbm$

Change in internal energy

$\Delta u= u_2-u_1$

= 38.09297-40.5485

= -2.4556

5 0

3 years ago

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m

Explanation:

see attached file below

3 0

3 years ago

What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a

Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18 )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0

3 years ago

Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C.

Maksim231197 [3]

Answer:

(d) all of the above

Explanation:

before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial

so option (D) will be correct because we need lubricate in all the given parts

8 0

3 years ago

Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.

levacccp [35]

Answer:

$\dot W_{out} = 3863.98\,kW$