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atroni [7]
1 year ago
10

10. A spring whose composition is not completely known might be either bronze (sp gr 8.8) or brass (sp gr 8.4). It has a mass of

1.26 g when measured in air and 1.11 g in water. Which is it made of? ​
Physics
1 answer:
frez [133]1 year ago
7 0

The element used is of bronze.

Now before we find out how we got the answer as bronze, firstly let us understand about the device we are talking about. That is, spring.

What exactly is a spring?

A spring is defined as a device that changes its shape in response to an external force, returning to its original shape when the force is removed. If we apply force or pressure in this device, it will change its shape but the moment we stop applying force, it will come back to its original state.

This means that this device is made of something flexible. So, the material it is made up of should also be flexible in nature.

How to find out flexibility of spring?

Flexibility of spring can be found as q = f Q, where f is the spring flexibility.

Here, in this question, we are given the mass of element in air and water. This shows a slight comparison between the mass of that particular element in air and water and thus, we get to know it is trying to draw our attention towards density.

Here, we have mass in air and water, thus, we can determine the density.

From these numbers, we can know that the density given is of bronze.

Hence, the element used is Bronze.

To know more spring aggregate, refer : brainly.com/question/20215636

#SPJ4

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A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV.
Mrrafil [7]

Answer:

The magnetic field required required for the beam not to be deflected  is B = 0.0036T

Explanation:

From the question we are told that

    The charge on the particle is q = +2e

    The mass of the particle is  m = 6.64 *10^{-27} kg

    The potential difference is V_a  = 1.8 kV = 1.8 *10^{3} V

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The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam  after the region having a potential difference of 1.8kV

               KE_b  =  PE_b

Generelly

              KE_b = \frac{1}{2} m v^2

And      PE_b = q V_a

 Equating this two formulas

              \frac{1}{2} mv^2 = q V_a

making v the subject

           v = \sqrt{\frac{q V_a}{2 m} }

Substituting value  

           v = \sqrt{\frac{ 2* 1.602 *10^{-19}  * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }

           v = 41.65*10^4 m/s

Generally the electric field between the plates is mathematically represented as

                 E = \frac{V_b}{d}  

Substituting value  

                 E = \frac{120}{8*10^{-3}}              

                E = 15 *10^3 NC^{-1}

the magnetic field  is mathematically evaluate    

                     B = \frac{E}{v}

                   B = \frac{15 *10^{3}}{41.65 *10^4}

                    B = 0.0036T

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