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hram777 [196]
3 years ago
15

A 3.06kg stone is dropped from a height of 10.0m and strikes the ground with a velocity of 7.00m/s. What average force of air fr

iction acts on it as it falls?
Physics
1 answer:
xxTIMURxx [149]3 years ago
3 0
<span>22.5 newtons. First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as: E = 0.5mv^2 where E = Energy m = mass v = velocity Substituting known values and solving gives: E = 0.5 3.06 kg (7 m/s)^2 E = 1.53 kg 49 m^2/s^2 E = 74.97 kg*m^2/s^2 Now ignoring air resistance, how much energy should the rock have had? We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So 3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2 So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is 299.88 J - 74.97 J = 224.91 J So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division. 224.91 J / 10 m = 224.91 kg*m^2/s^2 / 10 m = 22.491 kg*m/s^2 = 22.491 N Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
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Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

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In order to minimize the cost let us denote the side of the square bottom be 'a'

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The cost is:

C = 2sh

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Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

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Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

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\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}

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<h3>Answer:</h3><h3>we can say that:-</h3>
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