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natta225 [31]
3 years ago
6

Two forces, one of which double the other has resultant of 280N. if the direction of the large force is reversed and the remaini

ng unaltered, the resultant reduces to 160N. determine the magnitude of the force and the angle between the forces.
A body weighing 210N is resting on a horizontal table.A pull of 70 applied at an angle of 21 with the horizontal just causes the body to slide over the table. calculate the normal reaction and the coefficient of friction.​
Engineering
1 answer:
White raven [17]3 years ago
4 0

Answer:

point is 16 N.If their resultant is normal to the smaller force and has a magnitude of 8 N. Then two forces are.

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You can assume there is no pressure drop between the exit of the compressor and the entrance of the turbine. All the power from
Eddi Din [679]

Answer:

s6rt5x11j4fgu

j4

cf53yhu5

y4

hh

Explanation:

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6 0
3 years ago
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 =  0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = \frac{m}{\rho}

V = Av = \frac{\pi}{4} * d^2 * v1

V = \frac{\pi}{4} * 0.28^2 * 5

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = \frac{V}{v} = \frac{0.30787}{0.11418}

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = \frac{v2}{A2}

v2 = \frac{0.3705}{\frac{\pi}{4} * 0.28^2}

v2 = 6.017 m/s

7 0
3 years ago
A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of
lyudmila [28]

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 150 kPa

P_2 = final pressure of gas  = ?

V_1 = initial volume of gas   = v L

V_2 = final volume of gas  = \frac{v}{3}L

150\times v=P_2\times \frac{v}{3}  

P_2=450kPa

Therefore, the new pressure of the gas will be 450 kPa.

7 0
3 years ago
The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)
BartSMP [9]

Answer:

......................

Explanation:

3 0
3 years ago
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
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