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eduard
3 years ago
9

If a ping pong ball and a golf ball are both moving in the same direction with the same amount of kinetic energy, the speed of t

he ping pong ball must be
A. less than golf ball
B. same as golf ball
C. more than golf ball
D. impossible to know
Physics
1 answer:
Liono4ka [1.6K]3 years ago
5 0

If the kinetic energy of each ball is equal to that of the other,
then

(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²

Multiply each side by 2:

      (mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²

Divide each side by (mass of gb) and by (speed of ppb)² :

     (mass of ppb)/(mass of gb)  =  (speed of gb)²/(speed of ppb)²

Take square root of each side:

       √ (ratio of their masses)  =  ( 1 / ratio of their speeds)²

By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written.  Let's just come in from the
cold, and say it the clear, easy way:

If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.

If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.

The pingpong ball is moving faster than the golf ball.

The directions of their motions are irrelevant.

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Answer:

a=0.6\ m/s^2

Explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B, a=0.8\ m/s^2

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

m=\dfrac{F}{a}

m=\dfrac{100\ N}{0.8\ m/s^2}

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

(F-F')=ma

(100-25)=125\times a

a=\dfrac{75}{125}=0.6\ m/s^2

So, the new acceleration of the block is 0.6\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
How do you make a magnet with wire and electricity ?
AVprozaik [17]
<h2>Answer:</h2>

<u>By wrapping the wire along a solenoid and connecting it to electricity</u>

<h2>Explanation:</h2>

If you wrap a copper wire into coils and run an electrical current through it, you will create a magnetic field. If you rotate a permanent magnet as opposed to an item that has been magnetized inside a coil of copper wire, you can create an electrical current. The strength of magnetic field generated is proportional to the amount of current through the winding.

5 0
3 years ago
An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po
viktelen [127]

Answer:

1.232\times 10^{-5}\ T.

Explanation:

<u>Given:</u>

  • Current through the wire, passing through the origin, I_1 = 250\ A.
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According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to \mu_o times the net current threading the loop.

\oint \vec B \cdot d\vec l=\mu_o I.

In case of a circular loop, the directions of magnetic field and the line element d\vec l, both are along the tangent of the loop at that point, therefore, \vec B\cdot d\vec l = B\ dl.

\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\

\oint dl is the circumference of the Amperian loop = 2\pi r

Therefore,

B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_1 = 3.510\ m.

The magnetic field at the given point due to this wire is given by:

B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m,  concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_2 = 5.310\ m.

The magnetic field at the given point due to this wire is given by:

B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at r_r=-3.510\ m is given by

B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.

6 0
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In case of earth as light from the stars enters the earth atmosphere it bends towards smaller angle because the earth density increases as the light travel towards the earth troposphere from the exosphere as per the Snell's law described above.

b)Light rays that travel straight down do not bend, while rays that enter the Earth's atmosphere at a shallower angle get refracted and bend towards the normal, roughly following the direction of the Earth's curvature.

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