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guajiro [1.7K]
3 years ago
12

Suppose you have two magnets. Magnet A doesn't have its poles labeled, but Magnet B does have a clearly labeled north and south

pole. If these two are brought into contact with one another, which of the following could you expect? 
A. The side of Magnet A that's repelled by Magnet B's south pole must be Magnet A's north pole.

B. The side of Magnet A that's repelled by Magnet B's north pole must be Magnet A's south pole.

C. The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole.

D. The side of Magnet A that's attracted to Magnet B's north pole must be Magnet A's north pole.
Physics
2 answers:
Serggg [28]3 years ago
7 0
The opposite pole attracts each other and the similar pole repel each other. Which means only C is true.
Elina [12.6K]3 years ago
3 0

Before going to answer this question first we have to know the fundamental principle of magnetism.

A magnet have two poles .The important characteristic of a magnet is that like poles will repel each other while unlike poles will attract each other.

Through this concept the question can be answered  as explained below-

A-As per first option the side of  magnet A is repelled by the south pole  of magnet B. Hence the pole of a must be south .It can't be north as it will lead to attraction.

B-The side of magnet A is repelled by the  north pole of magnet B. Hence the side of A must be  north pole.It can't be a south pole.

C-The side of magnet A is attracted by the south pole  of magnet  B .Hence the side of magnet A must be north.Hence this is right

D-The side of magnet A is attracted by the north pole of magnet B. Hence the side of A must south.It can't be north as it will lead to repulsion.

Hence the option C is right.

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In a darkened room, a burning candle is placed 1.84 m from white wall. A lens is placed between candle and wall at a location th
KATRIN_1 [288]

Answer:

82.4 cm

Explanation:

The object and screen are kept fixed ie the distance between them is fixed and by displacing lens between them images are formed on the screen . In the first case let u be the object distance and v be the image distance

then ,

u + v = 184 cm

In the second case of image formation , v becomes u and u becomes v only then image formation in the second case is possible.

The difference between two object distance ie(  v - u ) is the distance by which lens is moved so

v - u = 82.4 cm

7 0
3 years ago
In his​ motorboat, Bill Ruhberg travels upstream at top speed to his favorite fishing​ spot, a distance of 120120 ​mi, in 33 hr.
photoshop1234 [79]

Answer:

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

Explanation:

x​ = the rate of the boat in still water

y​ = the rate of the current.

Distance travelled = 120 mi

Time taken upstream = 3 hr

Time taken downstream = 2.5 hr

Speed = Distance / Time

Speed upstream

\frac{120}{3}=x-y\\\Rightarrow 40=x-y

Speed downstream

\frac{120}{2.5}=x+y\\\Rightarrow 48=x+y

Adding both the equations

48+40=x-y+x+y\\\Rightarrow 88=2x\\\Rightarrow 44=x

40=44-y\\\Rightarrow 40-44=-y\\\Rightarrow y=4

The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>

8 0
3 years ago
Help plZ ASAP HELP PLZ
slamgirl [31]

Answer:

Waning Gibbous is correct

4 0
3 years ago
What properties of sound determine the volume of sound? Is this affected by the motion of the sound source?
Alik [6]
The amplitude determines the volume of the sound. This isn't affected by the motion of the sound source.
Hope this helps
6 0
3 years ago
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
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