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STatiana [176]
2 years ago
6

Mrs. Pham has 8 apples. She wants to give 3/4 of the

Engineering
2 answers:
Sergio039 [100]2 years ago
6 0
The answer would be 32/3 or if you want the simplified version it’s 10 2/3.
irakobra [83]2 years ago
6 0
6

2 = 1/4
4 = 2/4
6= 3/4
8=4/4
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Many households in developing countries prepare food over indoor cook stoves with no ducting system to exhaust the combustion pr
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Answer:

   C = 0.22857 ng / m³

Explanation:

Let's solve this problem for part the total time in the kitchen is

          t = 2h (60 min / 1h) = 120 min

The concentration (C) quantity of benzol pyrene is the initial quantity plus the quantity generated per area minus the quantity eliminated by the air flow. The amount removed can be calculated assuming that an amount of extra air that must be filled with the pollutant

amount generated

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         C = 0.2 + 0.01 120 / (40+ 2)

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3 years ago
Energy transfer in mechanical systems: During steady-state operation, a mechanical gearbox receives 70 KW of input power through
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Heat transfer rate(Q)= 1.197kW

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Three point bending is better than tensile for evaluating the strength of ceramics. a)-True b)- False
amid [387]

Answer:

a)-True

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3 years ago
A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

F_f= \mu N_2

where :

F_f = friction force

\mu = coefficient of friction

N_2 = normal friction

Given that:

\mu = 0.3

F_f = 0.3 N_2

Using the equation of equilibrium along horizontal direction.

\sum f_x = 0

P - F_f = 0

P = 0.3 N_2   ----- Equation (1)

To determine the moment about point B ; we have the expression

\sum M_B  = 0

0 = N_2*70-W*35-P*100

where;

P = horizontal force

N_2 = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = N_2*70-650*35-100*P

P = \frac{70 N_2-22750}{100} ----- equation (2)

Replacing  \frac{70 N_2-22750}{100}  for P in equation (1)

\frac{70N_2 -22750}{100} =0.3 N_2

N_2 = \frac{22750}{40}N_2 = 568.75 \ N

Plugging the value of N_2 = 568.75 \ N in equation (2)

P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100}  \\ \\ P = \frac{17062.5}{100}

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b)  From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

\sum M_A = 0

-(P × 100)+ (W×35) = 0

P = \frac{W*35}{100}

Replacing 650 N  for W

P = \frac{650*35}{100}

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

5 0
3 years ago
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