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3 years ago

Use the galvanometers to determine the amount and direction of the induced current. Which galvanometer is

Physics

1 answer:

docker41 [41]3 years ago

4 0

Answer:

Option B

Explanation:

Looking at the 3 galvanometer readings given above, for galvanometer A, the reading is -2 mA.

For galvanometer B, the reading is 4 mA.

While for galvanometer C, the reading is -5 MA

Thus, option B is correct.

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Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y

makkiz [27]

Answer:

<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:

1. Light waves are electromagnetic waves, while sound and water waves are mechanical waves, this is the first and principal difference.

2. Electromagnetic waves can propagate in vacuum (they do not need a medium or material), but mechanical waves obligatory need a material to propagate

3. Light waves are always transversal waves, this means the oscillatory movement is in a direction that is perpendicular to the propagation; but mechanical waves may be both: longitudinal waves (the oscillation occurs in the same direction as the propagation) or transversal waves.

4. Electromagnetic waves propagates at a constant velocity (Light velocity) while the velocity of mechanical waves will depend on the type of wave and the density of the medium or material.

5. Mechanical waves are characterized by the regular variation of a single magnitude, while electromagnetic waves are characterized by the variation of two magnitudes: the electric field and the magnetic field

6. Water waves are 2-dimensional waves, while the light and the sound are tridimensional spherical waves

7. Light waves transports energy in the form of radiation, while mechanical waves transport energy with material

3 0

3 years ago

REAL ANSWER = BRAINLIST LOOK AT PICS ATTACHED

inysia [295]

Answer:

there are no pictures attached.

5 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

Technician A says that the governor is used to measure vehicle speed in a hydraulically-controlled automatic transmission. Techn

igor_vitrenko [27]

Answer:

Both technicians are correct

Explanation:

In automobiles, governors are being used. Governors are also known for speed limiters as they can regulate and measure the speed of the automobile.

Hydraulic Transmission: Shifting is controlled by mechanical sensors according to vehicle speed and throttle position/manifold vacuum through valving operated by hydraulic 'pilot' signals.

Electronic Transmission: Shifting is controlled by an on-board computer that receives signals from electronic sensors.

3 0

3 years ago

Magma that cools and crystallizes on Earth's surface forms

Nostrana [21]

The correct answer is:

D. Extrusive rocks.

The explanation:

when extrusive igneous rocks form when magma reaches the Earth's surface a volcano and cools quickly. Most extrusive (volcanic) rocks have small crystals. Examples include basalt, rhyolite, andesite, and obsidian.

6 0

4 years ago

Read 2 more answers