How does the voltage applied to the electric motor affect the coil's number of rotations and its rotational direction

Physics

1 answer:

sleet_krkn [62]3 years ago

3 0

Higher voltage means faster rotation. Changing the voltage polarity changes the direction of the motor.

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Suppose an object is moving 2.1 m/s north on a river, but the river is flowing to the east at a velocity of 1.2 m/s. What is the

rewona [7]

Answer:

2.4 m/s

Explanation:

Given:

Velocity of the object moving north = 2.1 m/s

Velocity of the river moving eastward = 1.2 m/s

The resultant velocity is the vector sum of the velocities of object and river.

Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.

The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.

The magnitude of the resultant velocity (R) is given as:

$R^2=2.1^2+1.2^2\\\\R^2=4.41+1.44\\\\R=\sqrt{5.85}\\\\R=2.4\ m/s$

Therefore, the resultant velocity has a magnitude of 2.4 m/s.

8 0

3 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir

tia_tia [17]

Answer:

7560 Joules

Explanation:

$m_1$ = Mass of first car = $1.5\times 10^5\ kg$

$m_2$ = Mass of second car = $2\times 10^5\ kg$

$u_1$ = Initial Velocity of first car = 0.3 m/s

$u_2$ = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s$

Energy lost is

$\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J$