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AURORKA [14]
3 years ago
5

A long, straight, horizontal wire carries current toward the east. A proton moves toward the east alongside and just south of th

e wire. What is the direction of the magnetic force on the proton?
Physics
1 answer:
kondor19780726 [428]3 years ago
4 0

Answer:

north direction.

Explanation:

The wire carries a current towards the east . The magnetic field will make circular path around the wire in clockwise direction  so at a point just south of wire , magnetic field will be into the plane of paper containing wire. If we take east as x -axes , north as y axes then out of plane will form z axes. Hence direction of magnetic field will be -  z direction .

Magnetic field can be represented as - B k

Proton is moving towards east ie in + x direction so it can be represented as follows

velocity = V i

Force F = q( V i x -B k)

= ( BqV) j or + ve j direction or along north direction

So direction of force will be along north direction.

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The critical angle for water is 49°. If a ray of light
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Answer:

Snell's Law states

Ni sin i = Nr sin r

Judging from the question the source of the ray is in the water (directed up)

or NI = 1 / sin 49      Ni = 1.325 deg     the critical angle

From inside the pond:

Nr = 1.325 * sin 45 / 1 = 94 deg  

So refraction can occur  outside the pond and you do not have total internal refection.

 

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A pulley is most closely related to which simple machine?
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Explanation:

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Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm
Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

\eta\cdot \dot W = \dot K

\dot W = \frac{\dot K}{\eta} (Eq. 1)

Where:

\eta - Efficiency of fan, dimensionless.

\dot W - Electric power supplied fan, measured in watts.

\dot K - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2} (Eq. 2)

Where:

\rho_{a} - Density of air, measured in kilograms per cubic meter.

\dot V - Volume flow, measured in cubic meters per second.

A_{s} - Cross-sectional area of fan, measured in square meters.

If we know that \rho_{a} = 1.20\,\frac{kg}{m^{3}}, \dot V = 0.05\,\frac{m^{3}}{s}, \eta = 0.3 and A_{s} = 20\times 10^{-4}\,m^{2}, the power needed to be supplied by the fan is:

\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}

\dot K = 62.5\,W

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0
3 years ago
A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio
saveliy_v [14]

Answer: The coefficient of static friction is 3.85 and  The coefficient of kinetic friction is 2.8

Explanation:

in the attachment

6 0
3 years ago
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