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3 years ago

What would happen if you held the South Pole of one magnet near the North Pole of another magnet of the same size?

Physics

2 answers:

allsm [11]3 years ago

6 0

Answer:

(B) The magnets would be attracted to each other.

Explanation:

KiRa [710]3 years ago

4 0

The correct choice would be

(B) The magnets would be attracted to each other.

we know that like poles repel each other and unlike poles attract each other. hence when south pole of one magnet is placed near the north pole of other magnet, the two magnets attract each other due to interaction of magnetic fields of the two magnets. the magnetic field lines originating from north pole of one magnet ends at the south of other magnet

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Which of the following lists contains only examples of renewable sources of energy?

anyanavicka [17]

<h3><u>Answer;</u></h3>

A. wind, tidal, geothermal, and hydroelectric

<h3><u>Explanation</u>;</h3>

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3 years ago

A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$ is the final horizontal position of the projectile (the value we want to find)

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

$y$ is the final height of the projectile (the value we want to find)

$g=9.8m/s^{2}$ is the acceleration due gravity

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

$x=125m$ (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

$y=48.308m$ (6) This is the vertical final position of the projectile

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3 years ago

A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s

Natalka [10]

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

$KE_i = KE_f$