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3 years ago

Does groundwater flow or stay still?

Physics

1 answer:

zhenek [66]3 years ago

5 0

Answer:

yes it flows through flow paths.

Explanation:

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What is the strategy you use to solve word problems in physics?

Inga [223]

Answer:

The strategy we would like you to learn has five major steps: Focus the Problem, Physics Description, Plan a Solution, Execute the Plan, and Evaluate the Solution. Let's take a detailed look at each of these steps and then do an sample problem following the strategy.

3 0

2 years ago

A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm

Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

$v =\sqrt{\dfrac{T}{\mu}}$

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

$v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}$

v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

λ=2 L

λ=2 x 0.577 = 1.154 m

we now,

v = f λ

$f = \dfrac{v}{\lambda}$

$f = \dfrac{1074.49}{1.154}$

f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0

3 years ago

A group of students performed a compression experiment where they placed weights on top of a cylinder of material and measured t

kolbaska11 [484]

The material that the cylinder is made from is Butyl Rubber.

<h3>What is Young's modulus?</h3>

Young's modulus, or the modulus of elasticity in tension or compression, is a mechanical property that measures the tensile or compressive strength of a solid material when a force is applied to it.

<h3>Area of the cylinder</h3>

A = πr²

$A = \pi \times (0.02)^2 = 0.00126 \ m^2$

<h3>Young's modulus of the cylinder</h3>

$E = \frac{stress}{strain} \\\\E = \frac{F/A}{e/l} \\\\E = \frac{Fl}{Ae} \\\\$

Where;

e is extension

When 5 kg mass is applied, the extension = 10 cm - 9.61 cm = 0.39 cm = 0.0039 m.

$E = \frac{(5\times 9.8) \times 0.1}{0.00126 \times 0.0039} \\\\E = 9.97 \times 10^5 \ N/m^2\\\\E = 0.000997 \times 10^9 \ N/m^2\\\\E = 0.000997 \ GPa\\\\E \approx 0.001 \ GPa$

When the mass is 50 kg,

extension = 10 cm - 7.73 cm = 2.27 cm = 0.0227 m

$E = \frac{(50\times 9.8) \times 0.1}{0.00126 \times 0.0227} \\\\E = 1.7 \times 10^6 \ N/m^2\\\\E = 0.0017 \times 10^9 \ N/m^2\\\\E = 0.0017 \ GPa\\\\E \approx 0.002 \ GPa$