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Volgvan
3 years ago
12

REDDIT You look at a circular loop of wire such that the plane of the loop is perpendicular to your line of vision. The loop has

a constant current that is running through it clockwise. What will happen if a magnetic field were to be activated that is pointing in your direction
Physics
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

The circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.

Explanation:

Since the magnetic field, B points in my direction and the current, I is moving in a clockwise direction, the current is always perpendicular to the magnetic field and will thus experience a constant force, F = BILsinФ where Ф is the angle between B and L.

Since the magnetic field is in my direction, it is perpendicular to the plane of the circular loop and thus perpendicular to L where L = length of circular loop. Thus Ф = 90° and F = BILsin90° = BIL

According to Fleming's left-hand rule, the fore finger representing the magnetic field, the middle finger represent in the current and the thumb representing the direction of force on the circular loop.

At each point on the circular loop, the force is always directed towards the center of the loop and thus tends to compress it.

<u>So, the circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.</u>

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A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotatin
ycow [4]

Answer:

\frac{0.065}{r}

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

v=\sqrt{\mu gr}....................(1)

where \mu is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = 9.81m/s^2

r = ?

\mu=?

In order to solve for \mu, we can simply make it the subject of formula from equation (1) as follows;

v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

\mu=\frac{0.8^2}{9.81r}

\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}

8 0
3 years ago
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