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oee [108]
3 years ago
5

What is field selection process​

Physics
1 answer:
worty [1.4K]3 years ago
5 0

<h3>Answer:</h3>

It is a process of offering jobs to desired candidates and offering jobs to the desired applicant.

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True or false describing a metal as malleable means that it can be pounded into a new shape
vampirchik [111]
The answer is True.

Dictionary definition of <span>malleable. 
</span>
(of a metal or other material) able to be hammered or pressed permanently out of shape without breaking or cracking. 
3 0
3 years ago
"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after
wel

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

3 0
3 years ago
A 240 V circuit has 20 A flowing through it. How much power is it using?
Delvig [45]

Explanation:

P = IV

V = 240

I = 20

P = ?

P = 240 × 20

P = 4800 watts

6 0
3 years ago
Q12. How much force is applied on the body when 150 joule of work
ra1l [238]

Force = Work/distance

Force = 150/10

          =  15 Newtons

Force = 15 Newtons

Therefore, 15 newtons of force is applied to the body when 150 joules of work

is done in displacing the body through a distance of 10m in the  direction of the force.

4 0
3 years ago
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi
Keith_Richards [23]

Answer:

Explanation:

  a )

from lens makers formula

\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}

\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

 

5 0
3 years ago
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