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pogonyaev
3 years ago
5

As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.05 m 1

.05 m long rod as you jog at 3.27 m/s 3.27 m/s , holding the rod perpendicular to your direction of motion. What is the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.275 mV 0.275 mV across the rod? Express the answer in milliteslas.
Physics
1 answer:
S_A_V [24]3 years ago
3 0

Answer:

The strength of the magnetic field is 0.08 mT

Explanation:

Given:

Length of rod l = 1.05 m

Velocity of rod v  = 3.27 \frac{m}{s}

Induced emf \epsilon = 0.275 \times 10^{-3} V

According to the faraday's law

We know that the induced emf of rod is given by,

   \epsilon = Blv

Where B = magnetic field

For finding the magnetic field,

   B = \frac{\epsilon }{lv}

   B = \frac{0.275 \times 10^{-3} }{1.05 \times 3.27}

   B = 0.08 \times 10^{-3}

   B = 0.08 mT

Therefore, the strength of the magnetic field is 0.08 mT

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A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring
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Answer:

The spring stretched by x = 13.7 cm

Explanation:

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Mass = 3 kg

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Angle \theta = 34°

From the free body diagram

Force acting on the box = mg sin\theta

⇒ F = 3 × 9.81 × \sin34

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Since box is attached with the spring so a spring force also acts on the box.

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3 0
3 years ago
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t
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Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (\Delta t), in seconds:

\Delta t = t_{A}-t_{C} (1)

Where:

t_{C} - Time spent by the sound in concrete, in seconds.

t_{A} - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}

\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}}  \right)

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  } (2)

Where:

v_{C} - Speed of the sound in concrete, in meters per second.

v_{A} - Speed of the sound in the air, in meters per second.

x - Distance traveled by the sound, in meters.

If we know that \Delta t = 6.4\,s, v_{C} = 3000\,\frac{m}{s} and v_{A} = 343\,\frac{m}{s}, then the distance travelled by the sound is:

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  }

x = 2478.585\,m

The impact occured at a distance of 2478.585 meters from the person.

7 0
3 years ago
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