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stealth61 [152]
3 years ago
8

A student hangs a block from a light string that is attached to a massive pulley of unknown radius R, as shown in the figure. Th

e student allows the block to fall from rest to the floor. Which two of the following sets of data that could be measured or determined should the student use together to determine the final angular velocity of the pulley just before the block hits the floor? Select two answers. Justify your selections.
Physics
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

The mass of the block, the distance of the block above the floor, and the time it takes the block to reach the floor, because these quantities can be used to determine the acceleration of the block.

The radius and the mass of the pulley, because these quantities can be used together to determine the rotational inertia of the pulley.

Explanation:

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Name and describe three measures of central tendency used to summarize data.
OLEGan [10]

Answer:

Explanation:

You can think of it as the tendency of data to cluster around a middle value. In statistics, the three most common measures of central tendency are the mean, median, and mode. Each of these measures calculates the location of the central point using a different method.

There are three main measures of central tendency: the mode(the number that occurs most often) the median ( the middle value in the list of numbers) and the mean (the average number in the list of numbers).

5 0
3 years ago
Ignoring reflection at the air-water boundary, if the amplitude of a 10 GHz incident wave in air is 20 V/m at the water surface,
dimaraw [331]

Answer:

0.80267 m

Explanation:

E(z) = Electric field = 1 µV/m

E_0 = 20 V/m

z = Depth

\sigma = Conductivity = 0.1 S/m

\epsilon_r = 81

\mu = Impedance of free space = 120\pi\ \Omega

Frequency is given by

E(z)=E_0e^{-\alpha z}

Parameter is given by

\alpha=\dfrac{\sigma}{2}\sqrt{\dfrac{\mu}{\epsilon_r}}\\\Rightarrow \alpha=\dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{81}}\\\Rightarrow \alpha=20.94395\ N_p/m

From the first equation

1\times 10^{-6}=20e^{-20.94395z}\\\Rightarrow ln\dfrac{1\times 10^{-6}}{20}=-20.94395z\\\Rightarrow z=\dfrac{ln\dfrac{1\times 10^{-6}}{20}}{-20.94395}\\\Rightarrow z=0.80267\ m

The depth is 0.80267 m

7 0
3 years ago
Urgent<br> Please write the complete answer
Crazy boy [7]

Multiply field strength (N/kg) by mass (kg) to get weight (N)

At the start, the car is carrying

4.7 kg * (9.8 N/kg) = 46.06 N

of fuel.

At the end, it is carrying

3.0 kg * (9.8 N/kg) = 29.4 N.

Assuming the car remains completely intact, its weight was reduced by

46.06 N - 29.4 N = 16.66 N.

5 0
4 years ago
De ce nu putem masura o bandă​
icang [17]
I don’t speak Spanish sorry but I wish I could help you
3 0
3 years ago
Read 2 more answers
An electrical motor spins at a constant 2662.0 rpm. If the armature radius is 6.725 cm, what is the acceleration of the edge of
Gemiola [76]

Answer:

18.73 m/s^2

Explanation:

f = 2662 rpm = 2662 / 60 rps

r = 6.725 cm = 0.06725 m

Acceleration, a = r w

a = r x 2 x pi x F

a = 0.06725 × 2 × 3.14 × 2662 / 60

a = 18.73 m/s^2

3 0
3 years ago
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