Answer:
<em>a) 4.51 lbf-s^2/ft</em>
<em>b) 65.8 kg</em>
<em>c) 645 N</em>
<em>d) 23.8 lb</em>
<em>e) 65.8 kg</em>
<em></em>
Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound = 
slug
= 145/32.174 = <em>4.51 lbf-s^2/ft</em>
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds = kg
= 145/2.205 = <em>65.8 kg</em>
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = <em>645 N</em>
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
m/s^2 = 5.30 ft/s^2
= 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = <em>23.8 lb</em>
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = <em>65.8 kg</em>
Answer:
The correct option is;
c. Leaving the chuck key in the drill chuck
Explanation:
A Common safety issues with a drill press leaving the chuck key in the drill chuck
It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.
It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable
Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.
Answer:
radius = 9.1 × 
m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = ![\sqrt[3]{\frac{FL}{\sigma \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7BFL%7D%7B%5Csigma%20%5Cpi%7D%7D)
.................1
here F is applied load and is length
put here value and we get
radius = ![\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B5560%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B105%20%5Ctimes%2010%5E6%20%5Cpi%7D%7D)
solve it we get
radius = 9.1 × m
Answer:
(absolute).
Explanation:
Given that
Pressure ratio r
r=8
-----1
P₁(gauge) = 5.5 psig
We know that
Absolute pressure = Atmospheric pressure + Gauge pressure
Given that
Atmospheric pressure = 14.5 lbf/in²
P₁(abs) = 14.5 + 5.5 psia
P₁(abs) =20 psia
Now by putting the values in the above equation 1
Therefore the exit gas pressure will be 160 psia (absolute).
