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torisob [31]
2 years ago
11

A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance

it slides in 10seconds if the kinetic coefficient is 0.4.
​
Engineering
1 answer:
Talja [164]2 years ago
4 0

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

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Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, P_{t} = 8 kW = 8000 W

Now, the force, F is given by:

P_{t} = Force(F)\times velocity(v) = Fv               (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, F_{c} is provided by the gravitational force, F_{G}:

F_{c} = F_{G}

\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}

Thus from the above, velocity comes out to be:

v = \sqrt{\frac{GM_{e}}{R}}

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s

where

R = R_{e} + H

R = \sqrt{GM_{e}(\frac{T}{2\pi})^{2}}

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

8000 = F\times 3075.36

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3 years ago
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For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.
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Assumptions:

  • Steady state.
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Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

  • Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
  • Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

        r =\frac{V_1}{V_2}

  • Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
  • Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

         r = \frac{V_4}{V_3} = \frac{V_1}{V_2}

  • Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
  • Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

           \eta = 1-(\frac{1}{r^{\gamma - 1} } )

where:

           \gamma = \frac{C_{p} }{C_{v}} : specific heat ratio

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

           \gamma = 1.4

Answer:

See image attached.

5 0
3 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
3 years ago
The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inne
pshichka [43]

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

5 0
3 years ago
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