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AleksAgata [21]
2 years ago
7

Consider a loop of wire placed in a uniform magnetic field. Which factors affect the magnetic flux Φm through the loop?

Physics
1 answer:
dimaraw [331]2 years ago
4 0

Answer:

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

Explanation:

Magnetic flux is the scalar product of the magnetic field over the area

               Ф = ∫ B. dA

where B is the magnetic field and A is the area

Let's look at stationary, for which factors affect flow

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

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A box mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585N. The coefficient of kinetic frict
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the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

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2 years ago
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If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100
SCORPION-xisa [38]

Answer:

Charge enter a 0.100 mm length of the axon is 8.98\times 10^{-12} C

Explanation:

Electric field E at a point due to a point charge is given by

E=k \frac{q}{r^2}

where k is the constant =9.0 \times 10^9  Nm^2 / C^2

q is the magnitude of point charge and r is the distance from the point charge

Charges entering one meter of axon is 5.\times 10^{11} \times (+e)

Charges entering 0.100 mm of axon is 5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}

substituting the value of +e=1.6\times 10^{-19} C in above equation, we get charge enter a 0.100 mm length of the axon is

q=5.\times 10^{11} \times1.6\times 10^{-19}  \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C

3 0
2 years ago
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Nastasia [14]
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
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5 0
2 years ago
The total energy of a 0.050 kg object travelling at 0.70 c is:
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What is the unit c denotes here
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An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10
Marianna [84]

Answer:

a = 18.28 ft/s²

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given,

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mass of the object = 15 lb

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m = \dfrac{15}{32.174}\ slug

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now,

work done  is equal to change in kinetic energy

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now, acceleration of object

  a = \dfrac{v_f-v_o}{t}

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         a = 18.28 ft/s²

constant acceleration of the object is equal to 18.28 ft/s²

3 0
2 years ago
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