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3 years ago

Is the speed of the object during segment c greater than, equal to, or less than its speed during segment e?

1 answer:

4 0

This is a kinematic problem that asks us to determine the velocity of an object in a time frame given the plot of time in the x-axis and distance,x, in the y-axis. The figure that shows the behavior of the object is shown in the diagram. Segment C is going down which means the object is decelerating indicated by the negative slope. E on the other hand also has a negative slope as the line is also going down but less steep than that of c. slope is equal to change in y over the change in x. So given a definite change in y, a shorter change in x would make a greater slope, hence, becoming steeper. The speed then is C is higher than the speed in E.

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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-

maw [93]

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

4 0

3 years ago

A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 521.59 Hz

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.

8 0

3 years ago

A star is born when gas and dust from a nebula become so dense and hot that nuclear fusion starts. Which of the following forces

Kitty [74]

Your Question:
A star is born when gas and dust from a nebula become so dense and hot that nuclear fusion starts. Which of the following forces is responsible for the formation of a star?
A.
friction
B.
gravitation
C.
magnetism
D.
electromagnetic energy
Our Teams Answer: B. Gravitation

Our Teams aim's to please and We are happy to help! Its the answer is wrong please notify us quickly for we can fix it quickly! Thanks :)

-ExperimentsDIY

8 0

3 years ago

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Sedimentary rock turns into magnum through which process

alexandr402 [8]

Sedimentary rocks are formed when sediment is deposited out of air, ice, wind, gravity, or water flows carrying the particles in suspension. This sediment is often formed when weathering and erosion break down a rock into loose material in a source area.

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3 years ago

Which are characteristics of the image formed by an

Over [174]

Answer:

The image is real.

The image is inverted.

The image is bigger than the object.

Explanation:

I really don't know why... I got this question wrong and they said this was the answer. I wish I did. Sorry.

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3 years ago

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