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Vadim26 [7]
3 years ago
10

A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B f

ield (B is constant in time) and then back into a field free region to the right. The self inductance of the loop is negligible
True/False:

a. When leaving the field the coil experiences a magnetic force to the left.

b. Upon entering the field, a clockwise current flows in the loop.

c. Upon leaving the field, a clockwise current flows in the loop.

d. When entering the field the coil experiences a magnetic force to the right.
Physics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

A) False

B) False

C) True

D) False

Explanation:

A) False, because when leaving the field, the coil experiences a magnetic force to the right.

B) When the loop is entering the field, the magnetic flux through it will increase. Thus, induced magnetic field will try to decrease the magnetic flux i.e. the induced magnetic field will be opposite to the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is out of the plane of figure. Due to that reason, the current would be counterclockwise. So the statement is false.

C) When the loop is leaving the field, the magnetic flux through the loop will decrease. Thus, induced magnetic field will try to increase the magnetic flux i.e. the inducued magnetic field will be in the same direction as the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is also into the plane of figure. Due to that reason, the current would be clockwise. So the statement is true.

D) False because when entering the field magnetic force will be toward left side

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Morgarella [4.7K]

Answer:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Liquid P - Q = 3840\,J, Liquid Q - Q = 5500\,J, Liquid R - Q = 7800\,J, Liquid S - Q = 2856\,J

Explanation:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Let suppose that heat transfer rates between liquids and surroundings are stable. The quantity of the heat released is determined by the following expression:

Q = m\cdot c\cdot (T_{r} - T_{f}) (1)

Where:

m - Mass of the liquid, in kilograms.

c - Specific heat capacity, in joules per kilogram-degree Celsius.

T_{r} - Initial temperature of the sample, in degrees Celsius.

T_{f} - Freezing point, in degrees Celsius.

Liquid P (m = 1\,kg, c = 160\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 6\,^{\circ}C)

Q = (1\,kg)\cdot \left(160\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 6\,^{\circ}C)

Q = 3840\,J

Liquid Q (m = 1\,kg, c = 220\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 5\,^{\circ}C)

Q = (1\,kg)\cdot \left(220\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 5\,^{\circ}C)

Q = 5500\,J

Liquid R (m = 1\,kg, c = 300\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 4\,^{\circ}C)

Q = (1\,kg)\cdot \left(300\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 4\,^{\circ}C)

Q = 7800\,J

Liquid S (m = 1\,kg, c = 102\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 2\,^{\circ}C)

Q = (1\,kg)\cdot \left(102\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 2\,^{\circ}C)

Q = 2856\,J

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3 years ago
A skateboarder traveling with an initial velocity 9.0 meters per second,
meriva

Answer:

25m/s

Steps:

<em> First, The equation v= u + a * t shows us what we need to find, (the finale velocity). </em>

<em />

Second, we substitute the values given:

v= 9m/s + 4m/s2 * 4s

Last, We calculate the values:

Multiply 4m/s2 * 4s = 16m/s  

Add 9m/s + 16m/s

<u></u>

<u>Answer:  25m/s</u>

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Answer:

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