Answer:
A wheelbarrow, a bottle opener, and an oar are examples of second class levers
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 ×
= 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 ×
/ 3
so d = 0.0133 m
so diameter is 14 mm
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.
(At constant temperature and number of moles)
where,
= initial pressure of gas = 150 kPa
= final pressure of gas = ?
= initial volume of gas = v L
= final volume of gas =
Therefore, the new pressure of the gas will be 450 kPa.
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as


Integrating both sides we get

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get

Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of

Thus the remaining 125 meters will be covered with a constant speed of

in time equalling 
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds