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Sergio039 [100]
3 years ago
8

Ali was playing with his new U-shaped magnets which he attached with two trolleys and moved both of the trolleys in the same dir

ection so they can have a collision with each other but he wasn't successful. The car did not collide and he did not enjoy as he wanted to enjoy the collision between the trolleys. (a). With your understanding on magnetism explain why the trolleys did not collide with each other
Physics
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

The "trolleys" did not collide with each other because <em>their poles repelled. </em>

Explanation:

Magnets have poles located at their ends and these are the "north pole" and the "south pole."

If the north pole of one magnet will be faced to the south pole of another magnet, <em>they will attract each other</em> or <u>pull towards each other.</u> The opposite will happen when the south pole of one magnet will be faced with the same pole from another magnet. <em>They will repel each other</em> or<u> pull away from each other.</u>

Since the magnets attached to the trolleys were U-shaped, this means that the north and south poles are located at the<u> end of the magnets, which face the same direction. </u>So, once the magnets of the two trolleys move towards each other, their poles will repel because same poles will face each other.

So, this explains the answer.

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A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal,
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The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
J=F \Delta t
But the impulse is also equal to the change in momentum of the cannonball:
J=\Delta p
If we put the two equations together, we find
F \Delta t= \Delta p
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s
7 0
3 years ago
During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

5 0
3 years ago
What would decrease the resistance of wires carrying an electric current
Murljashka [212]
I believe the answer is A. shorter wires


3 0
2 years ago
Read 2 more answers
An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b
trapecia [35]

Answer: Option B: 1.3×10⁵ W

Explanation:

Power = \frac{Work \hspace{1mm} done}{Time}

P=\frac{W}{t}

Work Done, W= F.s

Where s is displacement in the direction of force and F is force.

\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

6 0
2 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
3 years ago
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