Answer:
Angular momentum of the system is 16221465.4617 kgm²/s
Explanation:
Given that;
length of the side of the triangle L = 82 M
m = 75.0 kg × 100 = 7500 kg
distance of each vertex from center R = L/√3 = 82/√3 = 47.34 m
effective acceleration a = 9.8 / 2 = 4.9 m/s²
we know that; effective acceleration is being provided by centripetal acceleration.
so
a = R × w²
rate of rotation w = √( a / R) = √( 4.9 / 47.34) = 0.3217 rad/seconds
Moment of Inertia I = 3mR²
we substitute
I = 3 × 7500 × (47.34)²
Also, Angular momentum L is expressed as;
L = I × w
so
L = 3 × 7500 × (47.34)² × 0.3217
L = 16221465.4617 kgm²/s
Therefore, Angular momentum of the system is 16221465.4617 kgm²/s
<span>The force that opposes the movement of an object through water is called drag. This is a type of frictional force. This force normally depends on the density and the viscosity of the fluid in question. The liquid which has more density and more viscosity or stickiness will produce a greater amount of drag force on an object than a fluid that is less dense and less viscous in nature. River water normally has less drag than that of sea water. </span><span> <span>
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Net force = (mass) · (acceleration)
= (69 kg) · (29 m/s²)
= (69 · 29) · (kg·m/s²)
= 2,001 Newtons upward
(about 450 pounds)
When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.
Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula

given that



now by using the above formula we will have



Answer:
Therefore the resistance of the conductor is 175Ω
Explanation:
Resistance:
- Resistance of a metallic conductor is directly proportional to its length(l).
- Resistance of a metallic conductor is inversely proportional to its cross section area(A).
The notation sign of resistance is R.
The unit of resistance is ohm (Ω).
Therefore,

and



ρ is the proportional constant.
It is also known as resistivity of that metal.
Given ρ=35×10⁻⁶Ω-m
l= 20 m
A= 4.0×10⁻⁶m²

=175Ω
Therefore the resistance of the conductor is 175Ω