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Blizzard [7]
3 years ago
13

A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct

ion) with respect to Earth, and one of the fragments is subsequently moving vertically with speed v, find the velocity ' of the other fragment immediately following the explosion. (Use the following as necessary: v.)
Physics
1 answer:
-Dominant- [34]3 years ago
4 0

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e  \dfrac{m}{2} .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v

The velocity of the other fragment immediately following the explosion is v .

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If two 100 ohms resistors are placed in series, their total resistance is what?
Sindrei [870]

Answer:

add the Resistance

2(100)= 200

2) 200 ohms

3 0
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Help ASAP <br> Just answer the first question for me please!
Usimov [2.4K]

Answer:

Im pretty sure its b

4 0
3 years ago
3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

3 0
3 years ago
Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction
kumpel [21]

Answer:

The  correct option is  H

Explanation:

From the question we are told that

    The index of refraction of  coating is  n_1

       The  index of refraction of material  is  n_2

   

Generally the condition for constructive for a thin film interference is mathematically represented

            2 *  t  = [ m  + \frac{1}{2}] \frac{\lambda}{n_1 }

Here  t represents the thickness

For minimum thickness  m =  0

So

           2 *  t  =0  + \frac{1}{2}\frac{\lambda}{n_1 }

=>        t  =\frac{\lambda}{4n_1 }

3 0
3 years ago
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