We use 1/o + 1/i = 1/f where o is the distance of the object, i as distance of the image and f is the focal length.
Substituting, <span>1/ 100 + 1 / i = - 1 /25 </span>
<span>i = - 20 cm </span>
<span>For the case of the problem,</span>
<span>o = (20 + 30) = 50 cm </span>
<span>f = 33.33. </span>Using 1<span> / i + 1 / o = 1/f , </span><span> </span><span>i = 100 cm </span>
<span>M = magnification = - i / o </span>
<span>m1 = -(-20)/100 = 20/100 = 0.2 </span>
<span>m2 = -100/50 = -2 </span>
<span>M = m1*m2 = -2 x 0.2 = -0.4.</span>
Explanation:
Given that,
Length of the spring, l = 50 cm = 0.5 m
Mass connected to the end, m = 330 g = 0.33 kg
The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm
(a) When the mass is connected, the force of gravity is balanced by the force in spring.
(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.
(c) The frequency of the oscillation is given by :
Hence, this is the required solution.
To lift the bag straight up takes (F · D) = (45 · 1.2) = 54 joules of energy (work).
Moving the bag horizontally 'across' gravity requires no work.
It doesn't matter how far.
Answer:
The answer is C, a thermometer reads 0.1 degree higher than it should.
