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2 years ago

Which of the following is a true statement about falling object safety?

Engineering

1 answer:

Anastaziya [24]2 years ago

6 0

Answer:

regregregrega

Explanation:

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Need help with both giving out brainlest for the people to help me

sammy [17]

The first one is d or the 4th answer choice and the second one is false. Hope this helps!

4 0

2 years ago

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a

Ratling [72]

Answer:

Explanation:

From the given question:

Using the distortion energy theory to determine the factors of safety FOS can be expressed by the relation:

$\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma y^2-\sigma x \sigma y+3 \tau_{xy^2}}$

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory can be expressed as:

$\tau_{max} = \dfrac{Syt}{2FOS}$

where;

$\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma y}{2})^2+ \tau _{xy^2$

a. Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}$

$\dfrac{350}{FOS}=160.35$

${FOS}=\dfrac{350}{160.35}$

FOS = 2.183

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2$

$\dfrac{350}{2 FOS} =88.51$

$\dfrac{350}{ FOS} =2 \times 88.51$

${ FOS} =\dfrac{350}{2 \times 88.51}$

FOS = 1.977

b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}$

$\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000$

$\dfrac{350}{FOS}=105.3565$

$FOS=\dfrac{350}{105.3565}$

FOS =3.322

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2$

$\dfrac{350}{2 FOS} ={ (\dfrac{110-100}{2})^2$

$\dfrac{350}{2 FOS} =25$

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}$

$\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}$

$\dfrac{350}{FOS}= 88.88$

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ (35)^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ 1225+ 400$

$\dfrac{350}{2 FOS} =40.31$

$FOS} =\dfrac{350}{2*40.31}$

FOS = 4.341

7 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve

lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa

The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0

2 years ago

Two resistors, A and B, individually connect to a 9V battery. A student notices that resistor A is warmer than resistor B. Which

dybincka [34]

Answer:

Resistor B

Explanation:

Since resistance is the opposition to the flow of current in a circuit,

first let assume the two resistors are connected in parallel to the voltage, recall that when connection is in parallel, the different amount of current pass through the resistors depending on the value with the small resistor having a lower resistance effect hence higher current will pass through

The energy dissipated in each resistor can be calculated as

$E=\frac{1}{2}IR^{2}t$ .

from the formula we can conclude that the energy value will be higher for the resistor with small resistance value. hence more heating effect which will cause it to be warm.

Also when connected individually the current flow from the voltage source will pass through the resistor which when we calculate the energy dissipated, the resistor with smaller value will be higher because it will draw more current which will in turn lead to a heating effect and cause the resistor to be warm. Hence we can conclude that the resistance B has greatest resistance value.

4 0

2 years ago