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3 years ago

How dense the medium is in the compression part of the wave, and how

1 answer:

4 0

How dense the medium is in the compression part of the wave and how rare the medium is in the rarefaction part of the wave is a measure of the longitudinal wave's amplitude.

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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri

mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a = - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0

3 years ago

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns

Tasya [4]

Complete Question

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.

The output from the secondary coil is 12 V

Answer:

The value is $N_s = 21 \ turns$

Explanation:

From the equation we are told that

The input voltage is $V_{in} = 120 \ V$

The number of turns of the primary coil is $N_p = 210 \ turn$

The output from the secondary is $V_o = 12V$

From the transformer equation

$\frac{N_p}{V_{in}} =\frac{N_s}{V_o}$

Here $N_s$ is the number of turns in the secondary coil

=> $N_s = \frac{N_p}{V_{in}} * V_s$

=> $N_s = \frac{210}{120} * 12$

=> $N_s = 21 \ turns$

4 0

4 years ago

12345 [234]

The similarity is that they both are types of bonds in molecules.
Ionic bonds are between a metal and a nonmetal.
Covalent bonds are between two nonmetals.

8 0

3 years ago

Samples of different materials, A and B, have the same mass, but the sample

madam [21]

Answer:

<u><em>A.)</em></u> The particles that make up material A have more mass than the

particles that make up material B.

HOPE that helps!!! :)

4 0

2 years ago

A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e

prohojiy [21]

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

$F_E$ is the electric force =?

The value of electric force will be;

$\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N$

Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

To learn more about the electrical force refer to the link;

brainly.com/question/1076352

7 0

2 years ago