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zheka24 [161]
3 years ago
5

The acceleration (in m/s^2) of a linear slider (undergoing rectilinear motion) within a If the machine can be expressed in terms

of its velocity (in m/s) and is given by α=1/4v^2 = if the slider starts at rest, find the time at which its velocity is 3.3 m/s.
Engineering
1 answer:
Inga [223]3 years ago
6 0

Answer:

47.91 sec

Explanation:

it is given that \alpha =\frac{1}{4v^{2}}

at t=0 velocity =0 ( as it is given that it is starting from rest )

we have to find time at which velocity will be 3.3 \frac{m}{sec^{2}}

we know that \alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}

4v^{2}dv=dt

integrating both side

\frac{4v^{3}}{3}=t+c---------------eqn 1

at t=o it is given that v=0 putting these value in eqn 1 c=0

so \frac{4v^{3}}{3}=t

when v=  3.3 \frac{m}{sec^{2}}

t=\frac{4}{3}\times 3.3^{3}

=47.91 sec

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A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the
myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm  ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²  

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If  τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I)   ⇒  V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒  Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0
3 years ago
A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th
ddd [48]

emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

Magnetic field (Initial)= 0.10 T

Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

=1000*3.14*0.5*10-4

=.0785

E=0.0785(.2/10*10-3)

=1.57 V

Thus, the emf generated is 1.57 V

4 0
3 years ago
A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f
KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) \frac{dm}{2}     ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = \frac{L}{\pi dm}      ...............2

here lead L = n × p

so tan(α) = \frac{2\times 2}{\pi 19}

α = 3.83°  

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and  now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) \frac{19\times 10^{-3}}{2}  

T = 1.59 Nm

so

power = \frac{2\pi N \ T }{60}     .................3

put here value

power = \frac{2\pi \times 300\times 1.59}{60}

power = 49.95 W

and

as φ > α

so it is self locking screw

 

8 0
3 years ago
Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.
saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 \mathbf{m^{3}}

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; v_1 = vg_1 = 0.0996 \mathbf{m^{3}} / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \mathbf{m^{3}} / kg

From temperature T₂ = 200⁰ C

v_f_2 = 0.0016 \ m^3/kg  

vg_2 = 0.127 \ m^3/kg  

Since  vf_2 < v_2 , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}

x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}

x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}

\mathsf{x_2 =0.78}

At temperature T₂, the specific internal energy u_f_2 = 850.6 \ kJ/kg , also ug_2 = 2594.3 \ kJ/kg

Thus,

u_2 = uf_2 + x_2 (ug_2 -uf_2)

u_2 =850.6  +0.78 (2594.3 -850.6)

u_2 =850.6  +1360.086

u_2 =2210.686 \ kJ/kg

<u>At state 3:</u>

Temperature T_3=T_2 = 200 ^0 C ,

V_3 = 2V_1 = 0.06 \ m^3

Specific volume v_3 = 0.2  \ m^3/kg

Thus; vg_3 =vg_2 = 0.127 \ m^3/kg ,

SInce v_3 > vg_3, therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at v_3 = 0.2  \ m^3/kg and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \ m^3/kg

The specific internal energy u_3 at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

u_2-u_1

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

u_3-u_2

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0
3 years ago
Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
2 years ago
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