Using lens equation;
1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)
Substituting;
1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm
Therefore, the object should be place 99.23 cm from the lens.
Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
<h3>
Free body diagram</h3>
The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;
/ W2
Ф → Ff
↓W1
where;
- Ff is the frictional force resisting the down motion of the box
- W1 is the perpendicular component of the box weight = Wcos(33)
- W2 is the parallel component of the box weight = Wsin(33)
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
Learn more about free body diagram of inclined objects here: brainly.com/question/4176810
Answer:
Explanation:
From the given information:
Distance 
Speed of the comet 
At distance 
where;
mass of the sun = 
To find the speed 
:
Using the formula:
![E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}](https://tex.z-dn.net/?f=E_f%20%3D%20E_i%20%2B%200%20%5C%5C%20%5C%5C%20%20K_f%20%2B%20U_f%20%3D%20K_i%20%2B%20U_i%20%20%5C%5C%20%5C%5C%20%3D%20%5Cdfrac%7B1%7D%7B2%7DmV_f%5E2%20%2B%20%20%5Cdfrac%7B-GMm%7D%7Bd%5E2%7D%20%3D%20%20%5Cdfrac%7B1%7D%7B2%7DmV_i%5E2%2B%20%5Cdfrac%7B-GMm%7D%7Bd_i%7D%20%5C%5C%20%5C%5C%20V_f%20%3D%20%5Csqrt%7BV_i%5E2%20%2B%202%20GM%20%5CBig%20%5B%20%20%5Cdfrac%7B1%7D%7Bd_2%7D-%20%5Cdfrac%7B1%7D%7Bd_i%7D%5CBig%20%5D%7D)
![V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}](https://tex.z-dn.net/?f=V_f%20%3D%20%5Csqrt%7B%289.1%20%5Ctimes%2010%5E%7B4%7D%29%5E2%20%2B%202%20%286.67%5Ctimes%2010%5E%7B-11%7D%29%20%2A%281.98%20%2A%2010%5E%7B30%7D%20%29%20%5CBig%20%5B%20%20%5Cdfrac%7B1%7D%7B6%2A10%5E%7B12%7D%7D-%20%5Cdfrac%7B1%7D%7B4.8%2A10%5E%7B10%7D%7D%5CBig%20%5D%7D)
