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A force 25 N makes an angle of 30,45 and 75 degree with x,y and z axis. What should be the corresponding force vector?

White raven [17]

Answer:

F = (21.651N, 17.678N, 6.470N)

Explanation:

The components of a vector in R ^ 3 are given according to their cosine directors, therefore, if we name F the force of 25N and Fx, Fy and Fz the components of the force with respect to the X, Y and Z axes , respectively:

Fx = FCos30 = 25 (0.866) = 21.651

Fy = FCos45 = 25 (0.707) = 17.678

Fz = FCos75 = 25 (0.259) = 6.470

F = (21.651N, 17.678N, 6.470N)

3 0

3 years ago

Air (ideal gas) is contained in a cylinder/piston assembly at a pressure of 150 kPa and a temperature of 127°C. Assume that the

Stels [109]

Answer:

The process is not possible.

Explanation:

We know for ideal condition, the work done for isothermal process is

$W_{ideal}$ = $P_{1}.V_{1} ln\frac{V_{2}}{V_{1}}$

and for ideal gas, we know PV = mRT

Therefore, $W_{ideal}$ = mRT $ln\frac{V_{2}}{V_{1}}$

= mRT $ln\frac{P_{1}}{P_{2}}$

= 0.287 x 400 $ln\frac{150}{450}$

= -126.12 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 126.12 kJ/kg )

Now we know for adiabatic compression process

P $V^{\gamma }$ = C

We know $\frac{T_{2}}{T_{1}}=(\frac{P_{2}}{P_{1}})^{\frac{\gamma -1}{\gamma }}$

$T_{2}$ = 556 K

For adiabatic work done, $W_{adiabatic}$ = $\frac{P_{1}\times V_{1}-P_{2}\times V_{2}}{\gamma -1}$

= $\frac{mR(T_{1}-T_{2})}{\gamma -1}$

= $\frac{0.287(400-556)}{1.45 -1}$

= -99.49 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 99.49 kJ/kg )

We know that in isothermal process, work input to the compressor is minimum. But in the above adiabatic polytropic process, work input to the compressor is less than the work done in the isothermal process.

Thus the process is not possible.

7 0

4 years ago

When using bits to represent fractions of a number, can you create all possible fractions? Why or why not?

Strike441 [17]

Answer:

no

Explanation:

There are an infinite number of fractions. You cannot create all possible fractions with a finite number of bits.

5 0

4 years ago

In plane stress, a prismatic bar of constant cross-section has an infinite length. a) True b) False

Law Incorporation [45]

Answer:

Option b) False

Explanation:

The given statement for a prismatic bar having constant area of cross section

is not correct and hence False as the length of the bar is not infinite.

The length of the prismatic bar must be very large in comparison to the thickness and width of the bar but has finite length and is not infinite.

Therefore, the correct statement is:

"In plane stress, a prismatic bar with constant area of cross section has finite length."

8 0

3 years ago

A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of

SIZIF [17.4K]

Answer:

number of mole is 31342.36 moles

mass is 125.369 kg

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = $\frac{4}{3} \pi r^3$

V = $\frac{4}{3}* 3.142* 4.5^3$ = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n = 76350000/2436 = 31342.36 moles

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m = 125.369 kg

3 0

3 years ago