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cricket20 [7]
2 years ago
13

A motor has coils with a resistance of 10.0 ohms and is supplied by a voltage of V = 1.20 x 102 V. When the motor is running at

its maximum speed, the back emf is 70.0 V. Find the current in the coils (a) when the motor is first turned on and (b) when the motor has reached its maximum rotation rate.
Physics
1 answer:
Vanyuwa [196]2 years ago
8 0

Hi there!

a)
When the motor is first turned on, the coils are initially stationary. Thus, there is no change in magnetic flux and, consequently, no induced emf.

Therefore:
iR = \epsilon - \epsilon_{back}

Since there's no back emf:
iR = \epsilon

Solving for i using Ohm's Law:
i = \frac{\epsilon}{R}\\\\i = \frac{120}{10} = \boxed{12.0 A}

b)

We are given that at max speed, the back EMF is 70.0 V.

Using the same equation as above:
iR = \epsilon - \epsilon_{back}

Plugging in the values:
10i = 120 - 70 \\\\10i = 50 \\\\

Solving for current:
i = \frac{50}{10} = \boxed{ 5.0 A}

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Daltons theory was identified using
Margaret [11]
<h2>Answer with Explanation </h2>

Dalton’s theory can be classified by the following hypotheses:

1) All material was formed of particles, unbreakable and strong construction segments.

2) All particles of a given component are indistinguishable in volume and characteristics

3) Compounds are determined by a mixture of two or more distinct kinds of atoms.

4) Chemical responses appeared in the rearrangement of the reacting atoms.

This theory was to explain all matter in terms of atoms and their characteristics, the law of conservation of volume and the law of constant composition.

7 0
4 years ago
A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside
neonofarm [45]

Answer:

\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}, level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}

\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt}  = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}

By replacing all known variables:

\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0
4 years ago
What is the net force on a bag pulled down by gravity with a force of 18 newtons and pulled upward by a rope with a force of 18
VARVARA [1.3K]

Answer:

0 N.

Explanation:

Force: This can be defined as the product of mass and the acceleration of the body. The S.I unit of force is Newton (N).

The expression of net force when both force act in the different direction is given as

F' = W-F ........................ Equation 1

Where F' = Net force on the bag, W = gravitational force on the bag, F = Force acting upward on the bag

Given: W = 18 N, F = 18 N.

Substitute into equation 1

F' = 18-18

F' = 0 N.

Hence the net force = 0 N.

4 0
3 years ago
A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
4 years ago
Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 1515 grams and velocity of 77 centimeters pe
Alexus [3.1K]

Answer:

The second kinetic energy is 162 J.

Explanation:

Given that,

Mass, m_1=15\ g

Velocity, v_1=7\ cm/s

Kinetic energy, K_1=147\ ergs

Mass, m_2=10\ g

Velocity, v_2=9\ cm/s

We need to find kinetic energy K_2. Kinetic energy is given by :

K=\dfrac{1}{2}mv^2

So,

\dfrac{K_1}{K_2}=\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}\\\\K_2=\dfrac{K_1}{\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}}\\\\K_2=\dfrac{147}{\dfrac{15}{10}\times \dfrac{7^2}{9^2}}\\\\K_2=162\ J

So, the second kinetic energy is 162 J.

4 0
3 years ago
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