Inventions for a 9 year old

The coefficient of static friction for both wedge surfaces is μw=0.4 and that between the 27-kg concrete block and the β=20° inc

balandron [24]

Assuming the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

<h3>Minimum value of force P</h3>

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°) + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

Learn more about Minimum value of force P here:brainly.com/question/20522149

7 0

2 years ago

Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?

zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

Th

To know more about Rankine cycle, visit: brainly.com/question/13040242

#SPJ4

4 0

1 year ago

A(n) ______ is used to measure fluid flow in engineering

Arte-miy333 [17]

Answer:

A pitot tube is used to measure fluid flow in engineering

3 0

2 years ago

A transmission line (TL) of length L and conductance per unit length G' is connected to an ideal constant voltage generator V. T

Andru [333]

Answer:

The Current will decrease by a factor of 2

Explanation:

Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.

Now, as the question says, the load is reduced to half its original value, we can write:

$P1 = \sqrt{3} (V) (I1) Cos\alpha ----- (1)$

$P2 = \sqrt{3} (V) (I2) Cos\alpha\\$

Since, P2 = P1/2,

$P1/2 = \sqrt{3} (V) (I2) Cos\alpha ----- (2)\\$

Dividing equations (1) and (2), we get,

P1 / (P1/2) = I1/ I2

$I2 = I1 / 2\\$

Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.

7 0

3 years ago

two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl

lesya692 [45]

The air flow necessary to remain at the lower explosive level is 4515. 04cfm

<h3>How to solve for the rate of air flow</h3>

First we have to find the rate of emission. This is solved as

2pints/1.5 x 1min

= 2/1.5x60

We have the following details

SG = 0.71

LEL = 1.9%

B = 10% = 0.1 a constant

The molecular weight is given as 74.12

Then we would have Q as

403*100*0.2222 / 74.12 * 0.71 * 0.1

= Q = 4515. 04

Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm

Read more on the rate of air flow on brainly.com/question/13289839

#SPJ1

7 0

2 years ago

Inventions for a 9 year old​

Inventions for a 9 year old