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3 years ago

There are two questions about SolidWorks.

Engineering

1 answer:

Nataliya [291]3 years ago

4 0

Answer:

1. It is a good practice to fully define a sketch to avoid having erroneous dimensions on the faces of a solid, this avoids that when it is required to make an assembly with the drawn part appear assembly errors.

2. The 2D sketch should always be done on a plane, so solidworks would ask you to select a plane on which you want to make the sketch, on the other hand, if it is a 3D sketch, solidworks allows you to do it without the need for Select any plane.

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If 1 uF capacitor is fully charged with 120 V across it, how much energy is stored in it? (a) 7.2 kJ (b) 7.2 mJ (c) 0.12 mJ (d)

jeyben [28]

Answer:

(b) 7.2 mJ

Explanation:

ENERGY STORED IN CAPACITOR: the energy is stored in capacitor in electric field

which can be calculated by expressions

E= $\frac{1}{2}$ c $v{^2}$

= $\frac{1}{2}$ $10^{-6}$ $120^{2}$

=7200× $10^{-6}$ J

= 7.2× $10^{-3}$ J

=7.2 mJ

4 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago

In order to avoid a rollover, what is the highest degree incline one should mow on? 10-degree incline 5-degree incline 30-degree

ser-zykov [4K]

Answer: B: 20-degree incline

Explanation:

A tractor user should avoid slopes of more than 20 degrees in order to avoid rollovers

6 0

2 years ago

Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Figure, with an initia

Rashid [163]

The heat transferred to and the work produced by the steam during this process is 13781.618 kJ/kg

<h3>How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

= 13331.618 + 450

Q = 13781.618 kJ/kg

Learn more about heat on:

brainly.com/question/13439286

#SP1

6 0

1 year ago

A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome

Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

BE / BC = 45 / 60 = 0.75

Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

vec(EF) = < -15 , -110 , 30 >

unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

vec(AD) = < 48 , -12 , 36 >

unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

unit (AD) = <0.7845 , -0.19612 , 0.58835 >

M_AD = unit(AD) . ( E x T_EF)

$M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]$

M_AD = 1835.73 + 116.49528 - 593.0568

M_AD = 1359.17 lb-in

3 0

3 years ago