Answer:
Explanation:
24 - gauge wire , diameter = .51 mm .
Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m
R = ρ l / s
1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]
= 8.42 x 10⁻² ohm
= .084 ohm
B ) Current required through this wire
= 12 / .084 A
= 142.85 A
C )
Let required length be l
resistance = .084 l
2 = 12 / .084 l
l = 12 / (2 x .084)
= 71.42 m
I would say C i'm not 100% sure
From the options provided in the question, the measurement which is not an SI base unit is volume.
<h3>What is SI base unit?</h3>
This is referred to as the standard and fundamental unit of measurement of various quantities or variables which is defined arbitrarily and not by combinations of other units.
Volume is a quantity which is derived from the combination of lengths in a three-dimensional manner which is why the formula is length× breadth×height and the unit is cm³. This is gotten from the combination of the unit of length which is cm.
This is therefore the reason why volume was chosen as the most appropriate choice.
Read more about Volume here brainly.com/question/463363
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Answer:
The density of gold is of 18 grams per cm3.
Explanation:
The mass density of a homogeneous material expresses how much mass of that material is present in a given volume. Since the density of an object is obtained by dividing its mass by its volume, to obtain the density of gold, its 90 grams of mass must be divided by its 5 cm3 volume, performing the following calculation:
90/5 = X
18 = X
Thus, the density of gold is 18 grams per cm3.
Answer:
See the answers below.
Explanation:
The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.
![Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]](https://tex.z-dn.net/?f=Cost%3D0.350%5BkW%5D%2A12%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%2A30%5Bdays%5D%2A4.5%5B%5Cfrac%7BRs%7D%7BkW%2Ahr%7D%20%5D%3D567%5BRs%5D)
The fuse can be calculated by knowing the amperage.
where:
P = power = 350 [W]
V = voltage = 240 [V]
I = amperage [amp]
Now clearing I from the equation above:
![I=P/V\\I=350/240\\I=1.458[amp]](https://tex.z-dn.net/?f=I%3DP%2FV%5C%5CI%3D350%2F240%5C%5CI%3D1.458%5Bamp%5D)
The fuse should be larger than the current of the circuit, i.e. about 2 [amp]
