<span>Atoms with the same atomic number but different atomic mass are called:
<span>Isotopes</span>
</span>
Compute the ball's angular speed <em>v</em> :
<em>v</em> = (1 rev) / (2.3 s) • (2<em>π</em> • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s
Use this to find the magnitude of the radial acceleration <em>a</em> :
<em>a</em> = <em>v </em>²/<em>R</em>
where <em>R</em> is the radius of the circular path. We get
<em>a</em> = <em>v</em> ² / (180 cm) = <em>v</em> ² / (1.8 m) ≈ 13.43 m/s²
The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude
∑ <em>F</em> = <em>m</em> <em>a</em>
where <em>m</em> is the mass of the ball. So, if <em>t</em> denotes the magnitude of the tension force, then
<em>t</em> = (1.6 kg) (13.43 m/s²) ≈ 21 N
Answer:
The current in the second loop will stay constant
Explanation:
Since the induced emf in the second coil, ε due to the changing current i₁ in the first wire loop ε = -Mdi₁/dt where M = mutual inductance of the coils and di₁/dt = rate of change of current in the first coil = + 1 A/s (positive since it is clockwise)
Now ε = i₂R where i₂ = current in second wire loop and R = resistance of second wire loop.
So, i₂R = -Mdi₁/dt
i₂ = -Mdi₁/dt/R
Since di₁/dt = + 1 A/s,
i₂ = -Mdi₁/dt/R
i₂ = -M × + 1 A/s/R
i₂ = -M/R
Since M and R are constant, this implies that i₂ = constant
<u>So, the current in the second wire loop will stay constant.</u>
Answer:
a) 
b) 
c) 
,
Explanation:
From the question we are told that:
Mass 
Velocity 
Coefficient of Rolling Friction 
a)
Generally the equation for The Propulsion Force is mathematically given by
b)
Therefore Power Required at
c)
Generally the equation for Work-done is mathematically given by
Therefore
Efficiency
Since
Energy in one gallon of gas is
Therefore
