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nadezda [96]
3 years ago
7

Two waves on one string are described by the wave functions

Physics
1 answer:
alex41 [277]3 years ago
3 0

Answer:

a) y_{R} (1.00,1.00) = -0.546\,cm, b) y_{R} (1.00,0.50) = -4.311\,cm, c) y_{R} (0.50,0.00) = 2.278\,cm

Explanation:

The principle of superposition points out that:

y_{R}(x,t) = \Sigma^{n}_{i=1} y_{i} (x,t)

Then, the results are:

a) y_{R} (1.00, 1.00) = y_{1} (1.00,1.00) + y_{2} (1.00,1.00)

y_{R} (1.00,1.00) = -0.546\,cm

b) y_{R} (1.00, 0.50) = y_{1} (1.00,0.50) + y_{2} (1.00,0.50)

y_{R} (1.00,0.50) = -4.311\,cm

c) y_{R} (0.50, 0.00) = y_{1} (0.50,0.00) + y_{2} (0.50,0.00)

y_{R} (0.50,0.00) = 2.278\,cm

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Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

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1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

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Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr
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Answer:

reliability

accuracy

Explanation:

If a reading of a measurement is consistently the same then the measurement is reliable.

If a reading of measurement is close the actual value of the measurement then the reading is accurate.

Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.

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Which best describes the way a sound wave is sent trough the radio?
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A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=
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Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

U=\frac{1}{2}k (\Delta x)^2

where

k is the spring constant

\Delta x is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

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and the amount of energy required is

U=\frac{1}{2}k d^2

In the second case, the spring is compressed from x=0 to x=-d, so

\Delta x = -d -0 = -d

and the amount of energy required is

U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2

so we see that the amount of energy required is the same.

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