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Lera25 [3.4K]
3 years ago
14

What circuit that has only one path for the current to follow?

Physics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Answer:

Series circuit

Explanation:

By definition series circuit is a closed circuit in which the flow of electric current follows only one path.

In a simple electrical circuit (which is a series circuit) where electric cell is the source of power, each component is connected in such a way that there is only one pathway current flows from the positive terminal(+) round to the negative terminal(-) via the external circuit.

Current having only one pathway to flow means that when one of the components is removed, there will be no current flow.

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A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma
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Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

(m_c + m_b)v_b + m_pv_p = 0

where m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg are the mass of the child, the boat and the package, respectively. , v_p = 10m/s, v_b are the velocity of the package and the boat after throwing.

(24.6 + 39)v_b + 5.8*10 = 0

63.6v_b + 58 = 0

v_b = -58/63.6 = -0.912 m/s

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How fast (in rpm) must a centrifuge rotate if a particle 7.50 cm from the axis of rotation is to experience an acceleration of 1
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3 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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