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2 years ago

A 10 ohms resistor is powered by a 5-V battery. The current flowing through the source is:

Physics

1 answer:

mario62 [17]2 years ago

4 0

Resistance=R=10ohm
Voltage=V=5V
Current=I

Applying ohm's law

$\\ \sf\longmapsto \dfrac{V}{I}=R$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{5}{10}$

$\\ \sf\longmapsto I=0.5A$

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A ball is falling after rolling off a tall roof. The ball has

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C.
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3 years ago

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How can the motion of a charged particle be used to distinguish between a magnetic field and an electric field in a certain regi

Helen [10]

Answer:

The charged particle follows a spiral path in a magnetic field.

Explanation:

A charge immersed in a region with an electric field experiences a force that acts along the same direction of the electric field. In particular:

- The force has the same direction as the electric field if the charge is positive

- The force has the opposite direction as the electric field if the charge is negative

Therefore, a charge moving in an electric field is accelerated along the direction of the electric field.

On the other hand, a charge in motion in a region with a magnetic field experiences a force that acts perpendicular to the direction of the field. This means that a charge in motion in a magnetic field will acquire a circular motion in the plane perpendicular to the direction of the magnetic field.

As a result, if the particle has also a original motion outside this plane, its final motion will consist of:

- A uniform motion along that direction, +

- A circular motion along the plane perpendicular to the field

So, the resultant motion of the particle will be a spiral path. So the correct answer is

The charged particle follows a spiral path in a magnetic field.

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3 years ago

What is used to turn the pointer of a galvanometer?

Sever21 [200]

Answer:

b. an electric current is used to turn the pointer of a galvanometer

Explanation:

well , galvanometer is an important device which is used to detect an electric current through a conductor or circuit.

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3 years ago

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1. how are weather data used to make forecasts?

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4 years ago

As a part of a project to construct a Rube Goldberg machine, a student wishes to construct a spring-loaded cube launcher. For ex

Alchen [17]

Answer:

1. The work done on the cube during the time the cube is in contact with the spring is 0.8023705 J

2. The speed of the cube at the instant just before the sliding cube leaves the ramp is approximately 31.5 cm/s

Explanation:

The given parameters of the Rube Goldberg machine are;

The distance from the free end of the spring to the top of the ramp, d = 17.0 cm = 0.17 m

The mass of the small cube to be launched, m = 119.0 g = 0.119 kg

The spring constant of the spring, k = 461.0 N/m

The angle of elevation of the ramp to the horizontal, θ = 50.0°

The coefficient of static friction of the wood, $\mu_s$ = 0.590

The coefficient of dynamic friction of the wood, $\mu_k$ = 0.470

The velocity of the cube at the top of the ramp, v = 45.0 cm/s = 0.45 m/s

The amount by which the cube is compressed, x = 5.90 cm = 0.059 m

The work done on the cube during the time the cube is in contact with the spring = The energy of the spring, E = (1/2)·k·x²

∴ E = (1/2) × 461.0 N/m × (0.059 m)² = 0.8023705 J

The work done on the cube during the time the cube is in contact with the spring= E = 0.8023705 J

2. The frictional force, $F_f$ = $\mu_k$ ·m·g·cos(θ)

∴ $F_f$ = 0.470 × 0.119 × 9.8 × cos(50) ≈ 0.35232 N

The work loss to friction, W = $F_f$ × d

∴ W = 0.35232 N × 0.17 m ≈ 0.05989 J

The work lost to friction, W ≈ 0.05989 J

The potential energy of the cube at the top of the ramp, P.E. = m·g·h

∴ P.E. = 0.119 kg × 9.8 m/s² × 0.17 m × sin(50°) ≈ 0.151871375 J

By conservation of energy principle, the Kinetic Energy of the cube at the top of the ramp, K.E. = E - W - P.E.

∴ K.E. = 0.8023705 J - 0.05989 J - 0.151871375 J ≈ 0.590609125 J

K.E. = (1/2)·m·v²

Where;

v = The speed of the cube at the instant just before the sliding cube leaves the ramp

∴ K.E. = (1/2) × 0.119 kg × v² ≈ 0.590609125 J

v² ≈ 0.590609125 J/((1/2) × 0.119 kg) ≈ 9.92620378 m²/s²

v = √(9.92620378 m²/s²) ≈ 3.15058785 m/s ≈ 31.5 cm/s

The speed of the cube at the instant just before the sliding cube leaves the ramp, v ≈ 31.5 cm/s.

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3 years ago