Explanation:
The three-second rule is recommended for passenger vehicles during ideal road and weather conditions. Slow down and increase your following distance even more during adverse weather conditions or when visibility is reduced. Also increase your following distance if you are driving a larger vehicle or towing a trailer.
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
= 0.75 < 25.84°
attached below is the remaining part of the solution
<u>B) Find the input current on the primary side in real units </u>
load current in primary = 31.38 < 25.84 A
<u>C) find the input power factor </u>
power factor = 0.9323 leading
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<em>attached below is the detailed solution </em>
Answer:
Because IC microcomputers are smaller and more versatile than previous control mechanisms, they allow the equipment to respond to a wider range of input and produce a wider range of output. They can also be reprogrammed without having to redesign the control circuitry
Explanation:
Answer:
See explanation
Explanation:
The complete question is
Write a SELECT statement that returns the same result set as this SELECT statement, but don't use a
join. Instead, use a subquery in a WHERE clause that uses the IN keyword.
SELECT DISTINCT category_name
FROM categories c JOIN products p
ON c.category_id = p.category_id
ORDER BY category_name
The join clause gives us all the registers of both tables, for which the category_id of categories coincides to the category_id in the table products. That is, we can first select the distinct category_id's from products and the see the categories from categories table, whose id is in that list. That means writing the following query:
Select distinct category_name
from categories c
where c.category_id in (
select distinct p.category_id
from products p)
order by category_name
Answer:
15.4 g/cm³, 17.4 g/cm³
Explanation:
The densities can be calculated using the formula below
ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)
fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8
a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³
b) density after infiltration with silver
fraction occupied by silver = 20 / 100 = 0.2
density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³