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3 years ago

From which end (north or south) of a bar magnet do magnetic field lines emerge?

1 answer:

5 0

They emerge from the north end. In modern magnetics we don't use the idea of north and south anymore but instead use a vector called the magnetization. It points toward what is traditionally called "north".

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As a 3.0 kg bucket is being lowered into a 10 m deepwell, starting from top, the tension in the rope is 9.8 N. theacceleration o

777dan777 [17]

Answer:

A) 6.5 m/s²

Explanation:

Mass of the bucket, m = 3.0 kg

depth of the well, d = 10 m

tension on the rope, T = 9.8 N

The net downward force on the bucket is given as;

T = mg - ma

where;

a is downward acceleration of the bucket

9.8 = (3 x 9.8) - 3a

9.8 = 29.4 - 3a

3a = 29.4 - 9.8

3a = 19.6

a = 19.6 / 3

a = 6.53 m/s² downwards

Therefore, the acceleration of the bucket is 6.53 m/s² downwards

8 0

3 years ago

Adding energy or increasing the speed of the particles at very low pressure in a solid would cause

weqwewe [10]

D. sublimation is correct

7 0

3 years ago

A mass m at the end of a spring vibrates with a frequency

Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

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

And given that ω = 2πf

the relation between frequency and mass becomes

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$