<h2>
<em>Answer:</em></h2><h2>
<em>Regular </em><em>object</em></h2>
- <em>Those </em><em>substance </em><em>which </em><em>have </em><em>fixed </em><em>geometrical </em><em>shape </em><em>are </em><em>called </em><em>regular </em><em>object.</em>
- <em>For </em><em>example</em><em>:</em><em> </em><em>books,</em><em>pencils,</em><em> </em><em>basketball</em><em> </em><em>etc.</em>
<h2>
<em>Irregular </em><em>object</em></h2>
- <em>Those </em><em>substance </em><em>which </em><em>do </em><em>not </em><em>have </em><em>geometrical</em><em> </em><em>shape </em><em>are </em><em>called </em><em>irregular</em><em> </em><em>object</em><em>.</em>
- <em>For </em><em>example:</em><em> </em><em>a </em><em>piece </em><em>of </em><em>stone,</em><em>a </em><em>broken </em><em>piece </em><em>of </em><em>brick,</em><em>leaf </em><em>etc.</em>
<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
Part a)

Part b)

Explanation:
Part a)
Force on the object due to spring force is given as

here we know that


so we have


Part b)
Acceleration of the object is given as



Answer: polar ice reflecting the Sun's light back toward space
A circuit with an impedance of 5 ohms and a voltage of 100 volts has a current flow of 20 A.
The expression of an electronic component, circuit, or system's resistance to alternating and/or direct electric current is called impedance, indicated by the letter Z. Resistance and reactance are two distinct scalar (one-dimensional) phenomena that combine to form impedance, a vector (two-dimensional) variable.
=
(for a pure resistor where Z=R,preserving the form of the DC Ohm's law).


I = 20 A
Therefore, A circuit with an impedance of 5 ohms and a voltage of 100 volts has a current flow of 20 A.
Learn more about impedance here;
brainly.com/question/2263607
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Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
= magnitude of current
magnitude of current is given as


= 1.87 A