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2 years ago

9

Straw polls are___. A. scientifically based B. representative samples C. biased sampling D. cluster samples

1 answer:

soldi70 [24.7K]2 years ago

8 0

Straw polls are considered to be a form of biased sampling. An example of a straw poll is an unofficial vote like who you would like to be the President without actually casting your ballot.

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Which of the following describes the mass of an object precisely ? *

timurjin [86]

Answer:

2.345 would be the most precious because you have more numbers to work with and exact numbers

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3 years ago

What are the chemical symbols for carbon aluminium hydrogen oxygen and sodium<br>

Elis [28]

El unico que se es el del oxígeno: o2

4 0

3 years ago

Problem 3 A person is standing in a room. Air temperature is 20 C. Assuming that his surface temperature is 30 C, calculate the

PilotLPTM [1.2K]

Answer:

the heat flux between the person and the room = 6.0036 x 10 17 w/m2

Explanation:

The appropriate formular for the steffan boltzmann constant is used as it relates to the heat flux and the steps is as shown in the attachment.

6 0

3 years ago

Two out-of-tune flutes play the same note. One produces a tone that has a frequency of 248 Hz, while the other produces 288 Hz.

zaharov [31]

Answer:

F = 268 Hz

Explanation:

The beat frequency is given as:

$Beat Frequency = |Frequency 1 - Frequency 2|\\$ |

So, for the first flute and tuning fork:

$20 Hz = |248 Hz - F|$

where,

F = Frequency of tuning fork

F = 248 Hz ± 20 Hz

F = 268 Hz (OR) 228 Hz

Now, for the second flute and tuning fork:

$20 Hz = |288 Hz - F|$

where,

F = Frequency of tuning fork

F = 288 Hz ± 20 Hz

F = 268 Hz (OR) 308 Hz

Since, 268 Hz is common from both calculations. Therefore, it will be the frequency of the tuning fork.

<u>F = 268 Hz</u>

3 0

2 years ago

The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

leonid [27]

Answer:

The magnitude of the force is $F_{net}= 1.837 *10^4N$

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

At t = 0 , $\theta = 20^o$

The rate at which the angle increases is $w = 2 \ ^o/s$

Converting this to revolution per second $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The length of the rope is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , The tension on the rope T = 18 kN

Mass of the para-sailor is $M_p = 75kg$

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

Now the derivative of displacement is velocity

So

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

represents the velocity, again the derivative of velocity gives us acceleration

So

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now to the time when the rope made angle of 30° with the water

generally angular velocity is mathematically represented as

$w = \frac{\Delta \theta}{\Delta t}$

Where $\theta$ is the angular displacement

Now considering the interval between $20^o \ to \ 30^o$ we have

$2 = \frac{30 -20 }{t -0}$

making t the subject

$t = \frac{10}{2}$

$= 5s$

Now at this time the displacement is

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity is

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

The linear acceleration is

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally radial acceleration is mathematically represented by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Generally angular acceleration is mathematically represented by

$\alpha_t = r \theta'' + 2 r' \theta '$

Now $\theta '' = \frac{d (0.0349)}{dt} = 0$

So

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The net resultant acceleration is mathematically represented as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the direction of the is acceleration is mathematically represented as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The force on the para-sailor along y-axis is mathematically represented as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The force on the para-sailor along x-axis is mathematically represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The net resultant force is mathematically evaluated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The direction of the force is

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

2 years ago