All categories

2 years ago

Bind hole, 38 diameter, .50 deep

Engineering

1 answer:

agasfer [191]2 years ago

4 0

Answer:

59.69021

Explanation:

38/.5 x 3.14159

You might be interested in

A well-designed product will increase?

Colt1911 [192]

Answer:

true

Explanation:

A well designed product will increase in sells and in stock.

8 0

2 years ago

Reduce the force F ij = + (2 5 ) kN to point A(2m,3m) that acts on point B( 3m,5m) - .

Alexxx [7]

Given :

Force, $\vec{F}= (2\hat{i} + 5\hat{j})\ kN$ .

Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )

To Find :

The work done by force F .

Solution :

Displacement vector between point A and B is :

$\vec{d} = (3-2)\hat{i} + (5-3)\hat{j}\\\\\vec{d} = \hat{i} + 2\hat{j}$

Now, we know work done is given by :

$W = \vec{F}.\vec{d}\\\\W= (2\hat{i} + 5\hat{j}).(\hat{i}+\hat{2j})\\\\W = (2\times 1) +( 5\times 2) \ kJ\\\\W = 12 \ kJ$

W = 12000 J

Therefore, work done by force is 12000 J .

6 0

3 years ago

To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr

alukav5142 [94]

<u></u> $\ T_{c}$ has greater effect.

<u>Explanation</u>:

$\eta_{\max }=1-\frac{T_{c}}{T_{A}}$

$T_{c}\\$ = Temperature of cold reservoir

$T_{H}$ = Temperature of hot reservoir

when $T_{c}$ is decreased by 't',

$\eta_{\text {incre }}$ = $1-\frac{\left(\tau_{c}-t\right)}{T_{H}}$

$=n \ + \frac{t}{T_{n}}$ $-(i)$

when ${T_{H}}$ is increased by 'T'

$\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)$

$\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}$

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

A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

Furkat [3]

Answer:

1700kJ/h.K

944.4kJ/h.R

944.4kJ/h.°F

Explanation:

Conversions for different temperature units are below:

1K = 1°C + 273K

1R = T(K) * 1.8

= (1°C + 273) * 1.8

1°F = (1°C * 1.8) + 32

Q/delta T = 1700kJ/h.°C

T (K) = 1700kJ/h.°C

= 1700kJ/K

T (R) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8R

= 944.4kJ/h.R

T (°F) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8°F

= 944.4kJ/h.°F

Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C

= 2°C

(273+5) - (273+3)

= 2 K

5 0

3 years ago