All categories

3 years ago

Which is true of magnetic field lines, but not electric field lines

Physics

2 answers:

Fed [463]3 years ago

8 0

Answer:

B) they show which way iron shavings would align themselves

Explanation:

Stels [109]3 years ago

7 0

Answer:

B) they show which way iron shavings would align themselves

Explanation:

Let's analyze each statement:

A) they are not affected by their own source --> this is true for both magnetic and electric fields. In fact, both fields are produced (and so affected) by a source (a magnet or a current in the magnetic field case, and an electric charge in the electric field case)

B) they show which way iron shavings would align themselves --> this is only true for the magnetic field. In fact, the pieces of iron will align according to the magnetic field; however, since they are electrically neutral, they are not affected at all by an electric field.

C) they re stronger near the source and get weaker farther away --> true for both magnetic and electric fields.

D) the closer the fields lines, the stronger the fields --> also true for both magnetic and electric fields.

So, the correct answer is B.

You might be interested in

Studies of gymnasts show that their high rate of injuries to the Achilles tendon is due to tensions in the tendon that typically

konstantin123 [22]

The horizontal components, being opposite in direction, will produce a shearing effect on the tendon, leading to injury. Thus, they will be added.
As the angle of 23 is from the vertical, we calculate the horizontal component by using
Weight acting * sin(23)
And multiply it by 2 to get the total shearing force
The weight acting is 12 times their body weight. Thus:
53 * 9.81 * 12 = 6239.2 Newtons
The total force acting on the tendon:
2 * sin(23) * 6239.2
= 4,880
= 4,900 Newtons (2 significant figures)

5 0

3 years ago

A child and sled with a combined mass of 50 kg, start from rest and slide down a frictionless hill that is 7.5 meters high. what

amid [387]

The mechanical enegia is the sum of the kinetic energy plus the potential energy
Kinetic energy = (1/2) * m * v ^ 2
Potential energy = m * g * h
Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
What is the mechanical energy of the sled at the top?
Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
Mechanical energy = (1/2) * (50) * (0) ^ 2 + (50) * (9.8) * (7.5) = 3675
Mechanical energy = 3675J
What is the mechanical energy of the sled at the bottom?
By conservation of energy we have that the energy in point 1 is equal to the energy in point 2
Mechanical energy = 3675J
What is the speed of the sled at the bottom of the hill?
Mechanical energy = 3675J = (1/2) * m * v ^ 2
clearing up v we have
(1/2) * (50) * v ^ 2 = 3675
v ^ 2 = 3675 * (2) * (1/50)
v = root (3675 * (2) * (1/50)) = 12.12 m / s
answer
3675J
3675J
12.12 m / s

3 0

3 years ago

Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light

Tpy6a [65]

Answer:

that technician A is right

Explanation:

The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.

From the above it is seen that technician A is right

4 0

3 years ago

A body moving in simple harmonic motion has an amplitude of 10cm and a frequency of 100Hz. Find i.mthe period of oscillation, ii

ohaa [14]

Explanation:

= 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

When displacement is y, then acceleration, a=−ω2y

Velocity, V=ωr2−y2

Case (a) When y=5cm=0.05m

a=−(10π)2×0.05=−5π2m/s2

V=10π×(0.05)2−(0.05)2=0

Case (b) When y=3cm=0.03m

a=−(10π)2×0.03=−3π2m/s2

V=10π×(0.05)2−(0.03)2=10π×0.04=0.4π m/s

Case (c) When y=0

a=−(10π)2×0=0

V=10π×(0.05)2−02=

hope it helps. pls follow and mark as brainliest

4 0

3 years ago

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/mN/m. At t=0 t=0 the block has velocity -

slega [8]

Answer:

The amplitude of the spring is 32.6 cm.

Explanation:

It is given that,

Mass of the block, m = 2 kg

Force constant of the spring, k = 300 N/m

At t = 0, the velocity of the block, v = -4 m/s

Displacement of the block, x = 0.2 mm = 0.0002 m

We need to find the amplitude of the spring. We know that the velocity in terms of amplitude and the angular velocity is given by :

$v=\omega\sqrt{A^2-x^2}$

$\omega=\sqrt{\dfrac{k}{m}}$

$\omega=\sqrt{\dfrac{300}{2}}$

$\omega=12.24\ rad/s$

So, $\dfrac{v^2}{\omega^2}+x^2=A^2$

$\dfrac{(-4)^2}{(12.24)^2}+(0.0002)^2=A^2$

A = 0.326 m

A = 32.6 cm

So, the amplitude of the spring is 32.6 cm. Hence, this is the required solution.

4 0

3 years ago