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3 years ago

A circuit in which two or more paths are connected to the voltage source

1 answer:

7 0

That is a parallel circuit and unlike a series circuit it has multiple paths so if a light bulb goes out it can still flow back to the power source. <span />

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Molds belong to a group of plants called <br> a.yeast<br> b.fungi<br> c.viruses

Marat540 [252]

B. fungi

I hope this helps!

5 0

4 years ago

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle

ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E = 5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0

3 years ago

When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0

adoni [48]

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁ = 48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s$

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

3 0

3 years ago

Please help !! In the diagram, q1 = +0.00200 C, q2 = 0.00180 C, and q3 = +0.00830 C. the net force on q2 is zero. how far is q2

VikaD [51]

Answer:

2.03715

Explanation:

$32364=8.99\cdot 10^9\cdot \frac{0.00180\cdot 0.00830}{2.03715^2}$

4 0

4 years ago

A second hand on a clock has a radius of 10 cm what is the circumference of a circle around which the secondhand travels

hram777 [196]

The answer has got to be 91

6 0

3 years ago