Ask question

Ask question

All categories

3 years ago

6

1. A drawing of a cabinet shows that its dimensions are 9cm. by 4cm. The drawing indicates 1:50 scaling. What are the actual dim

ensions of the cabinet?
2. Bongbong is tasked to draw a floor plan with a scale of 1:100. If the total length of the actual house is 8 meters, what will be its corresponding measurement in centimeters in floor plan?

3.The length of a bedroom measures 2.5 m., what will be the measurement in centimeters to be used in the drawing if it is in 1:50 scale?

1 answer:

yanalaym [24]3 years ago

5 0

Explanation:

This means that for every 1 cm on the drawing, there is 80 cm in reality. To put it another way, take this

1:80 means that the building is 80 times the size of the drawing

80:1 means that the drawing is 80 times the size of the building

If it were 80:1, the drawing itself would be over 100m long.

You might be interested in

All people<br><br><br>id 5603642259 pd 123456<br>on z o o m

algol [13]

Hello,

What is the ID Zoom for?

Like what in particular???

3 0

3 years ago

In the following scenario, what could the engineers have done to keep the bridge from collapsing?

yuradex [85]

Answer:

D

Explanation:

7 0

3 years ago

Name two types of battery chargers that are used in mechanics

tekilochka [14]

Primary batteries
Secondary batteries

7 0

3 years ago

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago

Define a separate subroutine for each of the following tasks respectively.

Valentin [98]

Answer:

I HAVE NO CLUEEEE

Explanation:

???????????????????/

5 0

3 years ago

Read 2 more answers