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DiKsa [7]
3 years ago
5

Two systems are in oscillation: a simple pendulum swinging back and forth through a very small angle and a block oscillating on

a spring. The block-spring system takes twice as much time as the pendulum to complete one oscillation. What could be done to make the two systems oscillate with the same period
Physics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

Here to make the two systems oscillate with same period either decrease the mass of the block system to one fourth or increase the length of the string to four times

Explanation:

we know that time period of block spring system is  2\pi\sqrt{\frac{m}{k} }

   where m = mass of the block

              k= spring constant

      and also time period of simple pendulum making small oscillations  is 2\pi\sqrt{\frac{l}{g} }

      where l = length of the pendulum    

                 g = acceleration due to gravity

   so here to make time period of the both systems to be same either we can decrease the block system time period to its half or double the time period of the pendulum.

 To decrease the time period of block spring system to half its value from above equation we need to decrease the mass of the block to one fourth

 and to increase the time period of the pendulum we need to increase length by four times .

   

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Four people return home from work and walk up the stairs to their own apartments in the same building which person is getting th
dmitriy555 [2]

Answer:

The person going to the highest apartment door.

Explanation:

Since gravitational potential energy, U = mgh where h is the height above the ground. The person with the highest gravitational potential energy would be the person going to the highest apartment door if we assume that their masses are the same.

6 0
4 years ago
Which of the following is true about inertia
Tamiku [17]

Answer:

The correct answers are It is the resistance of an object to changes in its motion, and It is a force

3 0
3 years ago
An object possessing an excess of 6.0x10^6 electrons has a net charge of
guajiro [1.7K]
A single electron has a charge of
e=-1.6 \cdot 10^{19}C
Therefore, if we have an excess of N=6.0\cdot10^6 electrons, the total net charge will be the product between the charge of a single electron and the total number of electrons in excess:
Q=Ne=(6.0\cdot10^6)(-1.6 \cdot 10^{-19}C)=9.6 \cdot 10^{-13}C
7 0
3 years ago
The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th
pickupchik [31]

Answer:

a

The focal length of the lens in water is  f_{water} = 262.68 cm

b

The focal length of the mirror in water is  f =79.0cm

Explanation:

From the question we are told that

    The index of refraction of the lens material = n_2

    The index of refraction of the medium surrounding the lens = n_1

 

The lens maker's formula is mathematically represented as

            \frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

Where f is the focal length

            n is the index of refraction

            R_1 and R_2 are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air  we have  

           \frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

    When immersed in liquid the formula becomes

          \frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2}  ]

The ratio of the focal length of the the two medium is mathematically evaluated as

           \frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }

From the question

      f_{air }= 79.0 cm

       n_2 = 1.55

and the refractive index of water(material surrounding the lens) has a constant value of  n_1 = 1.33

         \frac{f_{water}}{79}  = \frac{1.55- 1}{\frac{1.55}{1.44}  -1}

           f_{water} = 262.68 cm

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

   

7 0
3 years ago
A tugboat pulls a ship with a constant net horizontal force of 5.00 x 10 to the 3 power n and causes the ships to move through a
Mashutka [201]
Work is force times distance. So here we have
W=(5000N)x(3000m)=1.5x10^7J
Or 15MJ (megajoules)
5 0
3 years ago
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