A Stable Ride is the main characteristic of a catamaran hull.
What is catamaran hull?
Catamaran Hull is a multi-hulled watercraft, which have two parallel hulls of equal size.
The characteristic of a catamaran hull are as follows:
A catamaran hull has less volume than an equal size monohull.Catamaran hulls are distinguished by lighter displacement and shallower draft when compared to monohulls of the same length. Catamarans can be used in shallow waters as well. This feature makes catamarans quite flexible and versatile.Catamaran has multiple hulls, usually two parallel hulls of equal size. This geometric feature gives the craft an increased stability.Catamarans derives extra stability from its wide beam, in the place of a ballasted keel employed in a regular monohull sailboat.Therefore, Catamarans has a stable ride, according to their characteristic.
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Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q₁ and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.500 m. What is net force on 93?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
91
0.500 m-
-5.00 x 10-6 C
92
-5.00 x 10-6 C
93
0.500 m-
The answer to the question is 1.125 Newton
Formula for electrostatic force is F = ( K q1 q2 )/ r²
where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.
Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.
Electrostatic force between q1 and q3
r will be 0.500 + 0.500 = 1 m
F = ( K q1 q3 )/ r²
F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²
F₁₃ = 2.25 × 10⁻¹ N
Electrostatic force between q2 and q3
r will be 0.500 m
F = ( K q1 q3 )/ r²
F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²
F = (225 × 10⁻³) / (25 × 10⁻²)
F₂₃ = 9 × 10⁻¹ N
Total force on q3 will be F₁₃ + F₂₃
Total force on q3 = ( 2.25 × 10⁻¹ ) + (9 × 10⁻¹ ) N
Total force on q3 = ( 2.25 × 10⁻¹ ) + (9 × 10⁻¹ ) N
Total force on q3 = ( 11.25 × 10⁻¹ ) N
Total force on q3 = 1.125 N
Thus after solving we got the net force on q3 as 1.125 Newton
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