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1 year ago

what minimum volume must the slab have for a 42.0 kg woman to be able to stand on it without getting her feet wet?

1 answer:

8 0

0.525 m³ minimum volume must the slab have for a 42.0 kg woman to be able to stand on it without getting her feet wet.

<h3>How do you find the minimum volume?</h3>

Add the biggest possible inaccuracy to each metric, then multiply to get the biggest volume you can. Subtract the largest potential mistake of each measurement, then multiply, to determine the smallest volume that can be produced.

<h3>Briefing:</h3>

1000 kg/m³ = density of water

920 kg/m³ = density of ice

BF = Woman Weight + Slab Weight

Pvg = 42 + Pvg

V = 42 kg / (1000 - 920)

V = 42 / 80

V = 0.525 m³

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Answer:

a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

θ = 90 cos 90 = 0

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In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

θ = 180 cos 190 = -1

W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

m g sin θ L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀= + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

ΔU = -U₀

the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

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2 years ago

An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-

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Answer:

Power requirement P for the banner is found to be 30.62 W
Power requirement P for the solid flat plate is found to be 653.225 W
Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The Cd for banner was given, whereas the Cd for a flat plate is generally found to be around 1.28 which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
Power requirement P for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
The power can be determined this equation: P = F . v, where F represents the drag-force that we will need to determine and v represents the velocity of the airplane
The equation to determine drag-force is: $F = 1/2 * d * C_d * A$

In the drag-force equation Cd represents the co-efficient of drag and A represents the frontal area of the banner/plate/balloon (the object being towed)

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

Part a) We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = 1/2 * 0.06* 1.225 * 20 = 0.735 N. Now to find the power P we will use P = F . v i.e. 0.735 * 41.66 = 30.62 W

Part b) For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = 1/2 * 1.28 * 1.225 * 20 = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = 653.225 W

Part c) The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

Part d) The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : pi* r^2 (r = d / 2). Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = 40.08 W

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3 years ago

The noise floor, also known as additive white Gaussian noise (AWGN), is a continuous noise level that appears over a wide spectr

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Answer:

correct option is a. True

Explanation:

solution

the noise floor is AWGN ( additive white Gaussian noise )

and when viewed in the frequency domain, it is the continuous noise level

because as they have a uniform power over all the frequency.

so that it is additive white Gaussian noise

as we can say given statement is True

correct option a true

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