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2 years ago

When dropped from rest, any object will fall a total distance of 19.6 m in the first 2 seconds of its fall. How far

Physics

1 answer:

Setler79 [48]2 years ago

3 0

The distance the object willl fall durring the 2nd seconds is 4.9 m

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 2 - 1 = 1 s
Height (h) =?

<h3>How to determine the distance the object will fall during the 2nd seconds</h3>

The distance the object will fall during the 2nd seconds can be obtained as follow:

h = ut + ½gt²

h = (0 × 1) + (½ × 9.8 × 1²)

h = 0 + 4.9

h = 4.9 s

Thus, the distance the object will during the 2nd seconds is 4.9 m

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hovering mosquito is hit by a raindrop that is 45 times as massive and falling at 8.1 m/s , a typical raindrop speed. How fast i

Y_Kistochka [10]

Answer:

7.92 m/s

Explanation:

$m_1$ = Mass of raindrop = $45m_2$

$m_2$ = Mass of mosquito

$u_1$ = Initial Velocity of raindrop = 8.1 m/s

$u_2$ = Initial Velocity of mosquito = 0 m/s

$v$ = Velocity of center of mass

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{45m_2\times 8.1 + m_2\times 0}{45m_2 + m_2}\\\Rightarrow v=\frac{364.5m_2}{46m_2}\\\Rightarrow v=7.92\ m/s$

Hence, the velocity of the attached mosquito, falling immediately afterward is 7.92 m/s

4 0

3 years ago

Suppose that when you ride on your 7.50 kg bike the weight of you and the bike is supported equally by the two tires. If the gau

natali 33 [55]

Answer: 567 N

Explanation:

If the weight of the person and the bike are supported equally by the two tires, this means that the force acting on each tire, is half of the total weight.

We know that the gauge pressure in one tire, by definition, is equal to the force on the tire, over the contact surface between each tire and the road, so we can write:

P= F/A = (mrider g + mbike g) / A = (Fgrider + 7.5 kg. 9.8 m/s2) / 703 mm2 (1)

As the pressure data is given in lb/in2, it is needed to convert to N/mm2, as follows:

65.5 lb/in2 = 0.455 N/mm2.

Replacing in (1), and solving for Fgrider, we have:

Fgrider = 567 N

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2 years ago

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