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poizon [28]
3 years ago
15

Mike is conducting an experiment in which he places several different types of soil in a funnel and then pours one liter of wate

r through each. He is careful to pack equal volumes of the soil samples tightly in the funnel and to make sure that only water makes it out of the bottom. He then records the amount of time it takes for half of the water to pass through the soil and out the bottom of the funnel. He records his data in the table below.
Soil Sample Time (seconds)
Soil W 181
Soil X 27
Soil Y 119
Soil Z 50

Which soil sample is the least permeable?
A.
Soil W
B.
Soil Z
C.
Soil Y
D.
Soil X
Physics
2 answers:
Agata [3.3K]3 years ago
4 0

Answer:

soil W

Explanation:

Permeability is a measure of how easily water flows through soil. A soil that is made up of large, jagged rocks is likely to be more permeable than a soil that is made up of compact clay.  One way to test a soil's permeability is to pour water through it. More permeable soil will allow water to pass through it rapidly, while less permeable soil will block more of the water.

quester [9]3 years ago
3 0

Permeability is a measure of how fast a liquid can pass through a layer of solid. In this case, the lesser the time, the more permeable the solid is. Or the other way around, the bigger the time, the less permeable the solid is. Therefore the answer in this problem is:

<span> A. Soil W</span>

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Answer: The speed at the first quarter checkpoint is 0.74 m/s. The speed at the second quarter checkpoint is 1.40 m/s. The speed at the third quarter checkpoint is 1.61 m/s. The speed at the finish line is 1.89 m/s.

Explanation: I did the assignment and got it correct :)

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3 years ago
6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d
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Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0
3 years ago
As you approach a roundabout, you should __________________.
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The answer should be B
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The electric current running through the wire coil in an electric motor exerts force directly onto A) the battery. B) an aluminu
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<span>C)<span>a powerful magnet.

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3 years ago
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A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3
Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,  

\text { Electric power }(p)=\frac{V^{2}}{R}

\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}

\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

For calculating number of turns

\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

Given that,

80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }

\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}

\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .

We need to find the number of turns in the secondary winding \left(N_{S}\right) to run the bulb at 120W \left(P_{2}\right)

Firstly find the secondary voltage in the transformer use, \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}

V_{2}^{2}=\frac{120^{2} \times 120}{80}

V_{2}^{2}=\frac{1728000}{80}

V_{2}^{2}=21600

V_{2}=\sqrt{21600}

V_{2}=146.9 \mathrm{V}=V_{S}

Now, finding the number of turns in secondary coil. Use, \frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

\frac{30}{N_{S}}=\frac{65}{146.9}

N_{S}=\frac{30 \times 146.9}{65}

N_{S}=\frac{4407}{65}N_{S}=67.8

The number of turns in the secondary winding are 67.8 turns.

6 0
3 years ago
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