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2 years ago

Suppose the farthest distance a person can see without visual aid is 53 cm.

Physics

1 answer:

natita [175]2 years ago

8 0

Answer: a. 53cm b. Diverging c. 0.02dioptres

Explanation:

Since the person in question cannot see far object clearly, it shows that the individual is suffering from myopia (short sightedness).

If the object distance (u) is 53cm. for the person to be able to see far way, it means the image must be at infinity. This shows image distance v is infinity. Using the lens formula

1/f = 1/u+1/v

1/f = 1/53 + 1/Infinity

1/f = 1/53 + 0

f=53cm

b) The lens used is concave lens (diverging lens) to diverge all rays outwards

c) Power of a lens P will be 1/53.

P = 0.02dioptres

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Answer:

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$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

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