An average facility manager can build one new facility
during his or her career.
<span>A </span>facilities manager<span> is the ultimate organiser, making sure that a
workplace meets the needs of employees by managing all of the required
services. In this job, you will be responsible for the </span>management<span> of services and processes that support the core
business of an organisation.</span>
Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
Energy and Work have the same unit of measurement which is Joules in SI units.
Explanation:
- A Joule of Work is said to be done on an object when energy is transferred to that particular object.
- If two objects are involved, when one object transfers energy onto the second, a joule of work is said to be done by the first object.
- Work is also the application of force on an object over a distance. So Work = Force × Displacement
- Energy is neither created nor destroyed. It is in 2 forms - kinetic and potential.
- Kinetic energy is defined as the energy of a moving object while potential energy is known as the energy that is stored within an object.
- Kinetic Energy = 1/2 × mass × (velocity)²
- Potential Energy = mass × acceleration due to gravity × height
- Both energy and work are measured in Joules.
Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change
Answer:
Explanation:
The electric field outside the sphere is given as,
E = k Q /r²
here Q = n x 1.6 x 10⁻¹⁹ C
where n is the number of electons
if the dimeter of sphere d= 25 cm= 0.25 m
then the radius r = 0.125 m
we get
n= E r²/ k x 1.6 x 10⁻¹⁹ C
n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)
n = 14664731646
