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3 years ago

If a magnet could swing freely, what would happen to its north-seeking pole?

1 answer:

8 0

You clearly identified the pole you're talking about as the
"north-seeking" pole. Assuming your integrity and sincerity,
we would then naturally expect that pole to seek north, and
point to Earth's north magnetic pole.

I'm confident in this answer also because I have several of
these devices hanging from the ceiling of my office, and I can
attest to the fact that on most clear days, they do in fact point
toward Earth's north magnetic pole.

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Gre4nikov [31]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A 10 ohms resistor is powered by a 5-V battery. The current flowing<br> through the source is:

mario62 [17]

Resistance=R=10ohm
Voltage=V=5V
Current=I

Applying ohm's law

$\\ \sf\longmapsto \dfrac{V}{I}=R$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{5}{10}$

$\\ \sf\longmapsto I=0.5A$

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2 years ago

Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, re

ch4aika [34]

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of $62 lb / ft ^ 3$ (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

$W = \gamma A * \int_0^15 dy$

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

$W = (62)(14*7)\int^{15}_0 (15-y)dy$

$W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0$

$W = (14*7*62)[15(15)-\frac{(15)^2}{2}]$

$W = 683550ft-lbs$

Therefore the total work in the system is $683550ft-lbs$

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3 years ago

Look at the picture above and scientifically explain how eating a hamburger can cause energy transformation in three ways within

sergejj [24]

Answer:

CLAIM: When you eat the hamburger, your body uses it and breaks it down which is chemical energy than use that food for energy for mechanical energy which makes thermal energy

EVIDENCE: This happens every day to people across the world and also it is scientifically proven

REASONING: The reason why this evidence supports it is because as I said it happens everyday and when you feel energized after and apple after you were lazy that just happened

Explanation:

7 0

2 years ago

calculate the weight of an object of mass 100kg on the surface of a planet whose mass is 4.8×10^24 kg and diameter is 12000km.

notsponge [240]

Answer:

p= mv

where p is momentum

m is mass

v is velocity

so it's given p= 100kgm/sec

v= 4m/s

so putting in the formula

100= m × 4

m = 25kg

Explanation:

3 0

3 years ago