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2 years ago

An object is dropped from a height of 75.0 m above ground level. (a) determine the distance traveled during the first second.

Physics

1 answer:

barxatty [35]2 years ago

6 0

The relevant formula we can use in this case would be:

h = v0 t + 0.5 g t^2

where,

h = height or distance travelled

v0 = initial velocity = 0 since it was dropped

t = time = 1 seconds

g = 9.8 m/s^2

So calculating for height h:

h = 0 + 0.5 * 9.8 m/s^2 * (1 s)^2

<span>h = 4.9 meters</span>

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Answer:

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Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

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Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

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