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horsena [70]
3 years ago
13

Compasses line up with magnetic fields. A compass will line upA. Parallel to magnetic field lines, with the south pole pointing

in the direction of the field.B. Perpendicular to magnetic field lines.C. Parallel to magnetic field lines, with the north pole pointing in the direction of the field.D. Perpendicular to magnetic field lines, in a direction defined by the right-hand rule.
Physics
2 answers:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

C. Parallel to magnetic field lines, with the north pole pointing in the direction of the field.

Explanation:

When a magnet is placed in magnetic field then its North pole and south pole will experience magnetic force in opposite directions.

North pole will tend to move in the direction of magnetic field while south pole has tendency to move opposite to it.

Due to this opposite direction force magnet will experience torque if it is inclined at some angle with magnetic field

Due to this torque magnet will align in the direction of magnetic field such that its north pole will point in the direction of external field while south pole is align opposite to that

Olenka [21]3 years ago
4 0

you are welcome i hope god blesses you today

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10 points
enot [183]

Answer:

B

Explanation:

I just took the test

8 0
3 years ago
The magnitude of the voltage induced in a conductor moving through a stationary magnetic field depends on the _______ and the __
Virty [35]

The correct answer is: Option (D) length, speed

Explanation:

According to Faraday's Law of Induction:

ξ = Blv

Where,

ξ = Emf Induced

B = Magnetic Induction

l = Length of the conductor

v = Speed of the conductor.

As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed

3 0
3 years ago
Read 2 more answers
a slender rod of mass m and length l is released from rest in a horizontal position. what is the rod's angular velocity when it
Makovka662 [10]

Answer:

M g H / 2 = M g L / 2      initial potential energy of rod

I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2   kinetic energy attained by rod

M g L / 2 = 1/3 M L^2 * ω^2 / 2

g = 3 L ω^2

ω = (g / (3 L))^1/2

8 0
2 years ago
Two positive point charges repel each other with force 0.36 N when their separation is 1.5 m. What force do they exert on each o
egoroff_w [7]
The answer is: 0.81

I hope this helps :)
6 0
3 years ago
A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

7 0
3 years ago
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