Answer:
Its duration is 1.85*10⁻³ s or 1.85 ms
Explanation:
The intensity of electric current I is defined as the amount of electric charge Q (measured in Coulombs) that passes through a section of a conductor in each unit of time. The letter I is used to name the Intensity and its unit is the Ampere (A).
The intensity of electric current is expressed as:

where:
I: Intensity expressed in Amps (A)
Q: Electric charge expressed in Coulombs (C)
t: Time expressed in seconds (s)
Being:
Replacing:

Solving:
19500 A*t= 36 C

t= 1.85*10⁻³ s= 1.85 ms (being 1 s= 1,000 ms)
<u><em>Its duration is 1.85*10⁻³ s or 1.85 ms</em></u>
Answer:
Explanation:
a) Hardness is a measure of the resistance of a material to permanent deformation (plastic) on its surface,
Hardness tests play an important role in material testing, quality control and component acceptance.
Hardness test are needed to be perform as a <em>quality assurance procedure</em>, to validate materials are according to the specific hardness required,
We depend on the data to verify the quality of the components to determine if a material has the necessary properties for its intended use.
Through the years, the establishment of increasingly productive and effective means of testing, has given way to new cutting-edge methods that perform and interpret hardness tests more effectively than ever. The result is a greater capacity and dependence on "letting the instrument do the work", contributing to substantial increases in performance and consistency and continuing to make hardness tests very useful in industrial and R&D applications.
b)
- <u>Instrumental errors</u>: Instrument calibration is extremely important. An instrument with expired calibration may be generating erroneous data systematically.
- <u>Enviromental error: </u>An example is when surface preparation of the sample to be tested is poor, then the error can be presented when measuring the indentation on the sample to determine the hardness value.
Answer
The image formed by a concave mirror will be virtual, erect and magnified when the object is placed only between the pole and the principal focus of the mirror.
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Answer:
Maximum height of the ball, h(t) = 27.56 m
Explanation:
It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.
The height of the ball as a function of time t is given by :

h₀ is initial height, h₀ = 0
So,
.........(1)
For maximum/minimum height, 
...(2)
t = 1.31 s
Differentiating equation (2) wrt t
h''(t) = -32 < 0
So, at t = 1.31 seconds we will get the maximum height.
Put the value of t in equation (1)

h(t) = 27.56 m
Hence, this is the required solution.