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Ilya [14]
3 years ago
7

Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped

ends of two magnets together and they attract each other. Sally knows that opposite poles of magnets attract. She then brings the taped end of a third magnet (Magnet C) near each of the first two magnets. Predict her observations.
Physics
1 answer:
Artyom0805 [142]3 years ago
4 0
I already answered this quesiton. The fact is that there are only two kind of poles and since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles of the first two magnets are oppsosite.

Then, the taped pole of the third magnet has to be equal to one of the first two taped poles and opposite to the other of the first two taped poles.

That drives you to conclude (predict)  that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

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The object moves with BLANK from A to B. it BLANK from B to C. it moves with BLANK from C to D.
Ganezh [65]

Note that this is a position vs time graph.

From A to B, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

From B to C, the graph is a straight line with 0 slope. This indicates a constant position, i.e. the object remains stationary.

From C to D, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

4 0
2 years ago
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
loris [4]

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
3 years ago
A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?
Nostrana [21]

Answer:

The work done by friction was -4.5\times10^{5}\ J

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

W=\Delta KE

W=K.E_{f}-K.E_{i}

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2

Put the value of m and v

W=0-\dfrac{1}{2}\times1000\times(30)^2

W=-450000&#10;\ J

W=-4.5\times10^{5}\ J

Hence, The work done by friction was -4.5\times10^{5}\ J

3 0
3 years ago
A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic ener
Mademuasel [1]

The kinetic energy gained by the air molecules is 0.0437 J                

<h3 />

Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

P =mgh

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m

P = 0.0441 J

Finally, it will have kinetic energy. It is given by :

E= \frac{1}{2} mv^{2}

E= \frac{1}{2}×1.4 × 10⁻³ × (0.7)²

E = 0.000343 J

The  kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 -  0.0441

  = 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

Learn more about kinetic energy here:

brainly.com/question/8101588

#SPJ4

8 0
2 years ago
What is the approximate wavelength of a light whose first-order bright band forms a diffraction angle of 45.0° when it passes
sp2606 [1]
** Missing info: Lines per mm = 500 **

Ans: The wavelength is =  λ = 1414.21 nm

Explanation:
The formula for diffraction grading is:

dsinθ = mλ --- (1)

Where
d = 1/lines-per-meter = (1/500)*10^-3 = 2 * 10^-6
m = order = 1
λ = wavelength
θ = 45°

Plug in the values in (1):
(1) => 2*10^-6*sin(45°) = (1)λ
=> λ = 1414.21 nm
7 0
3 years ago
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