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poizon [28]
4 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]4 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]4 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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Which statements accurately describe Ernest Rutherford’s experiment? Check all that apply.
Dimas [21]

Answer:

Option (1), option (4) and option (5)

Explanation:

The main observations of Ernest Rutherford's experiment are given below:

1. most of the positively charged particles pass straight, it means there is an empty space in the atom.

2. Very few positively charged particles retraces their path.

So,

The positively charged particles were deflected because like charges repel, that means they are deflected by protons.

Almost all the positively charge concentrate in a very small part which is called nucleus.

7 0
3 years ago
Two small plastic spheres are given positive electrical charges. When they are 16.0 cm apart, the repulsive force between them h
Drupady [299]

Answer:

q = 7.542 x 10⁻⁷ C = 754.2 nC

Explanation:

The Coulomb's Law gives the magnitude of the force of attraction or repulsion between two charges:

F = kq₁q₂/r²

where,

F = Force of attraction or repulsion = 0.2 N

k = Coulomb's Constant = 9 x 10⁹ N m²/C²

r = distance between charges = 16 cm = 0.16 m

q₁ = magnitude of 1st charge

q₂ = magnitude of 2nd charge

Since, both charges are said to be equal here.

q₁ = q₂ = q

Therefore,

0.2 N = (9 x 10⁹ N m²/C²)q²/(0.16 m)²

(0.2 N)(0.16 m)²/(9 x 10⁹ N m²/C²) = q²

q = √(5.88 x 10⁻¹³ C²)

<u>q = 7.542 x 10⁻⁷ C = 754.2 nC</u>

7 0
3 years ago
If, in general, r were calculated as r =v/i, which circuit arrangement in part a of the experiment would have the smallest error
ladessa [460]

Concept: According to Ohm's Law, the flow of electric current through a conductor is directly proportional to the potential difference across it, provided physical conditions (like temperature, pressure, volume etc.) remains same.

v = ir

or, r = v / i

Here, current (i) is measured by Ammeter which should be connected in series of any electrical circuit.

voltage (v) is measured by Voltmeter which should be connected parallel to the external resistance (r).

In the given experiment, the first arrangement of the circuit will show the smallest error because the voltmeter is connected exactly parallel to the external resistance.

In the second arrangement, the voltmeter is connected across external resistance (r) and Ammeter (A) and in this case, the voltmeter will not measure the exact potential drop across the external resistance (r). So, there would be more error.

4 0
3 years ago
A circuit is made up of a moving magnet, a wire coil, a battery, and
Juli2301 [7.4K]
D should be the correct answer.
4 0
3 years ago
Read 2 more answers
A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I
Anuta_ua [19.1K]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = \frac{4\times \pi^2\times r^3}{GM} where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

8 0
3 years ago
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