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1 year ago

What direction would the north pole of a bar magnet point if you were to hang the bar magnet from a thin string?.

1 answer:

8 0

The end that is marked with ’N’ will point to the Arctic. The ‘S’ end will point toward the Antarctic. A bar magnet is ‘just’ a big heavy compass needle. The ‘N’ and ‘S’ marking are made to help travelers find their way around the globe.

Yes, it IS confusing to have the ’N’ end of a compass point toward the north axis of the earth. These conventions were well in place LONG before anyone understood that ‘Opposites Attract’. But now that we DO know that ‘Opposites Attract’ we are left with the unfortunate reality of compasses, bar magnets and the earth. There are gazillions of magnets and only one earth! So the ‘convention’ has been made that the magnets are labeled correctly! This means (unfortunately) that the Arctic area of the globe is actually the south end of the earthly magnet. The earth is a giant (but very weak) electromagnetic and the lines of the magnetic field point to the Arctic Regions which is why all out magnets point that way. The magnetic field line dive INTO the earth in the Arctic Regions and OUT of the earth in the Antarctic regions.

I haven’t used the word ‘pole’ here, because magnetic poles don’t actually exist (as far as we know). Every magnet is actually an electromagnetic with electrical charges in motion creating the magnetic field.

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An airplane is .68 Kilometers long. How many Millimeters long is the plane?

xxTIMURxx [149]

Answer:

680000

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4 0

2 years ago

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QUESTION 10

Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

$T_{1}sin(\theta)-T_{2}sin(\theta)=0$ (1)

Where:

T(1) is the tension due to the rope 1
T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0

2 years ago

A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds

horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

$F\Delta t = m\Delta v$

where:

F is the force exerted on the car

$\Delta t$ is the duration of the collision

m = 1400 kg is the mass of the car

$\Delta v=-27 m/s$ is the change in velocity of the car

We can re-write the equation as

$F=\frac{m\Delta v}{\Delta t}$

In the 1st collision, the time is 1.5 seconds, so the force is

$F_1=\frac{(1400)(-27)}{1.5}=-25,200 N$

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

$F_2=\frac{(1400)(-27)}{2.2}=-17,182 N$

Therefore, the force has been reduced by:

$F_2-F_1=-17,182-(-25,200)=8018 N$

4 0

3 years ago

Read 2 more answers

You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.

Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0

3 years ago

Two guitarists attempt to play the same note of wavelength 6.48 cmcm at the same time, but one of the instruments is slightly ou

k0ka [10]

The answer is λ₂ = 6.48 cm or 6.52 cm.

The out-of-tune guitar may have a wavelength between "6.48 cm" and "6.52 cm."

fb = |f2 − f1|

f₁ = 343/0.064

= 5276Hz

f₂ = 5276.9 Hz ± 17 Hz

f₂ = 5293.9 Hz or 5259.9 Hz

Now, calculating the possible wavelengths:

λ = 343/ 5259.9 or 343/ 5293.9

λ₂ = 6.48 cm or 6.52 cm

<h3>Why is beat frequency important?</h3>

When two waves with almost identical frequencies traveling in the same direction collide at a certain location, beats are produced. The opposing beneficial and harmful disruption causes the sound to alternatively be loud and weak whenever two sound waves with different frequencies reach your ear. This is referred to as beating.

The entire value of the frequency difference between the two waves is the beat frequency.

The following formula yields the beat frequency:

fb = |f2 − f1|

Learn more about beat frequency here:

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5 0

1 year ago

Read 2 more answers