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3 years ago

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An important news announcement is transmitted by radio waves to people who are 46 km away, sitting next to their radios, and by

sound waves to people sitting across the newsroom, 2.1 m from the newscaster. Take the speed of sound in air to be 348 m/s. What is the difference in time that the message is received

1 answer:

Sedbober [7]3 years ago

5 0

Answer:

$0.005847\ \text{s}$

Explanation:

Radio waves travel at the speed of light

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

$d_r$ = Distance between two radios = 46 km

$v$ = Speed of sound in air = 348 m/s

$d_s$ = Distance sound travels across the newsroom = 2.1 m

Time taken by the radio signal to reach the required location is

$t_r=\dfrac{d_r}{c}\\\Rightarrow t_r=\dfrac{46\times 10^3}{3\times 10^8}\\\Rightarrow t_r=0.000153\ \text{s}$

Time taken by sound to reach the required location is

$t_s=\dfrac{d_s}{v}\\\Rightarrow t_s=\dfrac{2.1}{348}\\\Rightarrow t_s=0.006\ \text{s}$

The time difference is

$t_s-t_r=0.006-0.000153=0.005847\ \text{s}$

The difference in time that the message is received is $0.005847\ \text{s}$ .

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A certain radio wave has a wavelength of 6.0 × 10-2m. What is its frequency in hertz?

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Answer:

The frequency of the wave is 5 x 10⁹ Hz

Explanation:

Given;

wavelength of the radio wave, λ = 6.0 × 10⁻²m

radio wave is an example of electromagnetic wave, and electromagnetic waves travel with speed of light, which is equal to 3 x 10⁸ m/s².

Applying wave equation;

V = F λ

where;

V is the speed of the wave

F is the frequency of the wave

λ is the wavelength

Make F the subject of the formula

F = V / λ

F = (3 x 10⁸) / (6.0 × 10⁻²)

F = 5 x 10⁹ Hz

Therefore, the frequency of the wave is 5 x 10⁹ Hz

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4 years ago

A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

m_a_m_a [10]

Answer:

a The kinetic energy is $KE = 0.0543 J$

b The height of the center of mass above that position is $h = 1.372 \ m$

Explanation:

From the question we are told that

The length of the rod is $L = 1.4m$

The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$

The angular speed at the lowest point is $w = 1.09 \ rad/s$

Generally moment of inertia of the rod about an axis that passes through its one end is

$I = \frac{mL^2}{3}$

Substituting values

$I = \frac{(0.140) (1.4)^2}{3}$

$I = 0.0915 \ kg \cdot m^2$

Generally the kinetic energy rod is mathematically represented as

$KE = \frac{1}{2} Iw^2$

$KE = \frac{1}{2} (0.0915) (1.09)^2$

$KE = 0.0543 J$

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

$KE = PE = mgh$

$0.0543 = mgh$

$h = \frac{0.0543}{9.8 * 0.140}$

$h = 1.372 \ m$

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3 years ago

Read 2 more answers

What two factors determine the density of a fluid?

tekilochka [14]

Answer:

Mass and volume.

Explanation:

The equation for density is always mass divided by volume. To determine the density of a fluid, you would need to find its volume and its mass.

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4 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

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Answer:

System management

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