Answer:
height ≈ 60.60 m
Explanation:
The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.
Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.
Using tangential ratio,
tan 22° = opposite/adjacent
tan 22° = h/150
h = 150 × tan 22°
h = 150 × 0.40402622583
h = 60.6039338753
height ≈ 60.60 m
Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline
Question:Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline . Which of them is correct ?
Answer: Technician B is correct
Explanation: Two types of engines exist , the two stroke (example, used in chainsaws) is a type of engine that uses two strokes--a compression stroke and a return stroke to produce power in a crankshaft combustion cycle and the four stroke engines(eg lawnmowers) which uses four strokes, 2-strokes during compression and exhaustion accompanied by 2 return strokes for each of the initial process to produce power in a combustion cycle.
While a 2 stroke system engine, requires mixing of oil and fuel to the crankshaft before forcing the mixture into the cylinder and do not require a pressurized system. The 4 stroke system uses a splash and pressurized system where oil is not mixed with gasoline but drawn from the sump and directed to the main moving parts of crankshaft through its channels.
We can therefore say that Technician A is wrong while Technician B is correct
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required
Answer:
To answer this question we assumed that the area units and the thickness units are given in inches.
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
[tex] V = A*t [/tex]
Where:
A: is the surface area = 160
t: is the thickness = 0.002
Assuming that the units given above are in inches we proceed to calculate the volume:
[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]
Now, using the density we can find the mass:
[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]
Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):
[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
2. The block is released from rest at the position shown, figure 1. The coefficient of
kinetic friction over length ab is 0.22, and over length bc is 0.16. Using the
principle of work and energy, find the velocity with which the block passes
position c.
Answer:
Velocity = 4.73 m/s.
Explanation:
Work done by friction is;
W_f = frictional force × displacement
So; W_f = Ff * Δs = (μF_n)*Δs
where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24
Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;
mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).
Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;
ΔKE2 + (0.16)(mg cos 24)(2).
Plugging in the relevant values, we have;
1.22mg = ΔKE1 + 0.603mg
ΔKE1 = 1.22mg - 0.603mg
ΔKE1 = 0.617mg
Also,
0.813mg = ΔKE2 + 0.292mg
ΔKE2 = 0.813mg - 0.292mg
ΔKE2 = 0.521mg
Now total increase in Kinetic Energy is ΔKE1 + ΔKE2
Thus,
Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg
Putting 9.81 for g to give;
Total increase in kinetic energy = 11.164m
Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m
m cancels out to give; ½v² = 11.164
v² = 2 × 11.164
v² = 22.328
v = √22.328
v = 4.73 m/s.