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ololo11 [35]
2 years ago
6

The uniform dresser has a weight of 90 lb and rests on a tile floor for which the coefficient of static friction is 0.25. If the

man pushes on it in the horizontal direction, thetatheta= 0o, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 150 lb, determine the smallest coefficient of static friction between his shoes and the floor so that he does not sli
Engineering
1 answer:
Juliette [100K]2 years ago
3 0

Answer:

a) F = 736.065\,lbf, b) \mu_{k} = 0.15

Explanation:

a) The uniform dresser is modelled by using the following equations of equilibrium:

\Sigma F_{x} = F - \mu_{k}\cdot N = 0

\Sigma F_{y} = N-m\cdot g=0

After some algebraic manipulation, the following formula is derived:

F = \mu_{k}\cdot m \cdot g

F = (0.25)\cdot (90\,lbm)\cdot (32.714\,\frac{ft}{s^{2}} )

F = 22.5\,lbf

b) The man is described by the following equations of equilibrium:

\Sigma F_{x} = -F + \mu_{k}\cdot N = 0

\Sigma F_{y} = N-m\cdot g=0

After some algebraic manipulation, the following formula for the static coefficient of friction is:

\mu_{k} = \frac{F}{m\cdot g}

\mu_{k} = \frac{22.5\,lbf}{150\,lbf}

\mu_{k} = 0.15

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(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
2 years ago
An electrical utility delivers 6.25E10 kWh of power to its customers in a year. What is the average power required during the ye
Sindrei [870]

Answer:

The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh

However, the average power P, required for a year, t  = ? Kw

Expressing their relationship, we will have

             U = P x t

Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)

          t = 8760 hours

P = \frac{62500000000}{8760}

P = 7134.7Kw

Hence, the average power required during the year is 7,135Kw

Now to calculate the energy used by the power plant in a year (in quads)?

Recall, Efficiency, η = Power Output/Power Input (100)

so, we have

η = P₀/P₁, given

0.45 = \frac{7134.7Kw}{P₁}

P₁ = 15,855Kw

the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules

So to convert this to quads, Note;

1 quads of energy = 10¹⁵ Joules

The total energy used is 0.000000139 quads

Now to find the cubic feet of natural gas required to generate this power?

Note: 0.29Kwh of Power generated  = 1 cubic feet of natural gas used

Since, the power plant generated = 62500000000Kwh

The cubic feet of natural gas used = \frac{62500000000}{0.29}

Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.

8 0
2 years ago
Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace
Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0
2 years ago
A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hou
Novosadov [1.4K]

Answer:

0.867

Explanation:

The driver population factor (f_{p})can be estimated using the equation below:

f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}

The value of the heavy vehicle factor (f_{HV}) is determined below:

The values of the E_{T} = 2 and E_{R} = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:

f_{HV} = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833

Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:

f_{p} = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867

6 0
2 years ago
Oliver is designing a new children’s slide to increase the speed at which a child can descend. His first design involved steel b
AVprozaik [17]

Answer:

The correct option is;

A) Steel becomes too hot in the Sun and can burn the children

Explanation:

The properties of steel includes;

Low specific heat capacity, high thermal and electrical toughness, high hardness, high tensile strength, high yield strength, appreciable elongation, high fatigue strength, can easily corrode, high malleability and ability to creep

Therefore, due to the low specific heat capacity, which is 0.511 J/(g·°C) and high conductivity of steel which is about 32 W/(m·k), the temperature of the steel can rapidly rise and the hot steel surface can readily conduct the heat, (due to the temperature difference) to other bodies that come in contact

8 0
2 years ago
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