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dolphi86 [110]
3 years ago
14

__________________ is the process by which sediments get pressed together.

Physics
2 answers:
Alekssandra [29.7K]3 years ago
8 0

Answer:

compaction

Explanation:

Shkiper50 [21]3 years ago
5 0
The answer is compaction
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The advantage of a fixed pulley on a flag pole is that it ___________.(1 point)
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Decreases the input force
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In Rutherford's experiment, which of the following proved that the
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Answer:

b) the alpha particles were found to be attracted to the nucleus

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A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea
Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
R_T = R_1 + R_2 + ... + R_n

In parallel:
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}

2.

We can use Ohm's law to solve for the current in the circuit.

i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

V = iR\\\\V = (6)(15) = \boxed{90 V}

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2 years ago
The force required to stretch a Hooke’s-law
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I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K


Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J


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3 years ago
What are two examples of common units for each of the above measurements
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Density: g/mL, kg/cubic meter  

Volume: L, teaspoon  

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