Answer:
1st – Place the film canister on the <u>scale</u>.
2nd – Slide the large <u>weight </u>to the right until the arm drops below the line and then move it back one notch.
3rd – Repeat this process with the <u>top</u> weight. When the arm moves below the line, back it up one groove.
4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.
5th – Add the amounts on each beam to find the total <u>mass </u>to the nearest tenth of a gram.
Explanation:
The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.
The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.
Answer:
471392.4 N
Explanation:
From the question,
Just before contact with the beam,
mgh = Fd.................... Equation 1
Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.
make f the subject of the equation
F = mgh/d................ Equation 2
Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m
Constant: g = 9.8 m/s²
Substitute into equation 2
F = 1900(4)(9.8)/0.158
F = 471392.4 N
Answer:
Service.
Explanation:
Service equipment and service conductors runs from the service point to the service disconnecting point. Service provides the delivery of electrical energy from it's production point to the consumer where it is meant to be utilized.
Answer:
P = 1000000[Pa] = 1000 [kPa]
Explanation:
To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

where:
P = pressure [Pa] (units of pascals)
F = force = 100 [N]
A = area = 100 [mm²]
But first we must convert the units from square millimeters to square meters.
![A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2} } =0.0001[m^{2} ]](https://tex.z-dn.net/?f=A%3D100%5Bmm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7D%20m%5E%7B2%7D%20%7D%7B1000%5E%7B2%7Dmm%5E%7B2%7D%20%20%7D%20%3D0.0001%5Bm%5E%7B2%7D%20%5D)
Now replacing:
![P=100/0.0001\\P=1000000[Pa]](https://tex.z-dn.net/?f=P%3D100%2F0.0001%5C%5CP%3D1000000%5BPa%5D)
Answer:
0.015%
Explanation:
Data provided in the question:
Length by which the measuring tape can be off, δL = 0.42 cm
Total measured length for which Error of δL is observed, L = 28 m
Now,
we know,
1 m = 100 cm
Thus,
28 m = 28 × 100 = 2800 cm
Percent uncertainty = [δL ÷ L] × 100%
= [0.42 ÷ 2800] × 100%
= 0.015%