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Elan Coil [88]
3 years ago
5

A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f

rom a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is B. Part A What is the magnitude Φinitial of the magnetic flux through one turn of the coil before it is rotated?
Physics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

\phi_i = BA

Explanation:

magnetic flux is the count of magnetic field lines passing through a given loop or area

As we know that magnetic flux is given by the formula

\phi = \vec B. \vec A

here we also know that magnetic field B and plane of the coil is perpendicular in initial position

So the area vector is always perpendicular to the plane of the coil

so the angle between magnetic field and area vector is parallel to each other and this angle would be zero

so magnetic flux of the coil initially we have

\phi = BAcos0 = BA

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What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to
agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy (Q), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components (U_{l,ice}, U_{l,steam}), in joules, and two sensible component (U_{s,w}), in joules, that is:

Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam} (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w}) (2)

Where:

m - Mass, in kilograms.

h_{f,w} - Latent heat of fussion of water, in joules per kilogram.

h_{v,w} - Latent heat of vaporization of water, in joules per kilogram.

c_{ice} - Specific heat of ice, in joules per kilogram per degree Celsius.

c_{w} - Specific heat of water, in joules per kilogram per degree Celsius.

\Delta T_{ice} - Change in temperature of ice, measured in degrees Celsius.

\Delta T_{w} - Change in temperature of water, measured in degrees Celsius.

If we know that m = 5\,kg, h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}, h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}, c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, \Delta T_{ice} = 10\,^{\circ}C and \Delta T_{w} = 100\,^{\circ}C, then the amount of thermal energy is:

Q = 15167500\,J

The amount of thermal energy needed is 15167500 joules.

7 0
3 years ago
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Kinetic energy is the energy possessed by an object when that object is moving in space. The higher the mass of an object or higher the speed of an object the higher the kinetic energy will be.

So to calculate the Kinetic Energy we can use the following formula

K.E=(1/2)*m*v^2

Inserting the values in formula gives:

K.E=1/2*7.26*2^2

14.52J

This is the final answer which gives the kinetic energy of the ball.

6 0
3 years ago
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MissTica

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

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