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3 years ago

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Which of the following factors influences the speed of a wave? I. the type of wave II. the amplitude of the wave III. the medium

in which the wave travels

2 answers:

Nata [24]3 years ago

3 0

I. the type of wave ... electromagnetic or mechanical

(II. amplitude doesn't influence the speed of any wave)

III. the medium in which the wave travels ... completely determines the wave speed

S_A_V [24]3 years ago

3 0

Answer: The correct answers are (I) and (III).

Explanation:

The expression for the relation between the speed of the wave and the frequency is as follows;

$v=f\lambda$

Here, f is the frequency and $\lambda$ is the wavelength.

The speed of the wave depends on the medium in which the wave travels.

It depends on the type of wave but it does not depend on the amplitude of the wave.

For example, the speed of the sound wave is more in the solid medium in comparison to the liquid and gaseous medium as the solid medium is more dense. Here, the particles are more closer.

The speed of light is more than the speed of the sound. Light is transverse wave whereas sound wave is longitudinal wave.

Therefore, the factors which influence the speed of a wave are as follows;

the type of wave (II )and (III) the medium in which the wave travels.

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A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 8.0 m/s2. A 20 kg mas

nadya68 [22]

Answer:

a) 60 N

b) 860 N

Explanation:

Given that,

$m_1$ = 100 kg

$m_2$ = 20 kg

$a_{1}$ = 8.0 $\frac{m}{s^{2}}$

$a_{2}$ = 3.0 $\frac{m}{s^{2}}$

a) By Newton's Law,

∑ $F_{m_2,x} = f_k = m_{2} * a_{2}$

∑ $F_{m_2,y} = F_{N} - m_{2} * g = 0$

Hence,

$f_k = m_2 * a_2 = 20 * 3 = 60 N$

b) By Newton's Law

∑ $F_{m_1,x} = F = m_{1} * a_{1}$

Hence,

$F = m_{1} * a_{1} = 100 * 8 = 800 N$

Net force acting on 100 kg mass,

$F_{net} = F + f_k = 800 + 60 = 860 N$

6 0

3 years ago

The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

Mademuasel [1]

Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

$(1). \: \:\beta =10 log(\dfrac{I}{I_0})$

The ratio of the intensities can be written as

$\frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} }$

$\dfrac{I}{I_0} = \dfrac{A_0}{A}.$

And since

$A = 4\pi r^2$

and

$A_0 = 4 \pi r_0^2$ ,

$\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2} = \dfrac{r_0^2}{r^2}$

meaning

$\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.$

Putting this into equation (1), we get:

$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

$100dB=10 log(\dfrac{r_02}{(5m)^2})$

$10= log(\dfrac{r_0^2}{25})$

by taking both sides to the exponent:

$10^{10}= \dfrac{r_0^2}{25}$

$r^2 = 25 *10^{10}\\r = 5 *10^5$

Now equation (2) becomes

$\beta = 10log(\dfrac{25*10^{10}}{r^2})$

when the intensity level is 85 dB we have

$85 = 10log(\dfrac{25*10^{10}}{r^2})$

$8.5 = log(\dfrac{25*10^{10}}{r^2})$

take both sides to exponents and we get:

$10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}$

$10^{8.5} =\dfrac{25*10^{10}}{r^2}$

$r^2 = \dfrac{25*10^{10}}{10^{8.5}}$

$\boxed{r = 28.11m}$

Thus, $28.11m$ far from the speaker the intensity drops to 85 dB.

7 0

3 years ago

If we take away a lot of energy from a liquid what phase will it change to ?

DanielleElmas [232]

The liquid would change to gas

8 0

3 years ago

A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s.If the car after this acceleration s

aalyn [17]

Answer:

The time it takes to stop is 13.75 seconds

Explanation:

A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;

v = u + a·t

Where;

v = The final velocity of the body

a = The acceleration of the body

t = The time of acceleration (accelerating period) of the body

u = The initial velocity of the body

The given parameters for the acceleration of the car are;

The initial velocity of the car, u = 0 m/s (a car starting from rest)

The constant acceleration of the car, a = 5.5 m/s²

The acceleration duration, t = 6 s

Therefore, we have;

The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s

The final velocity of the car after the acceleration, v = 33 m/s

When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'

The given parameters when the car slows down are;

The deceleration, -a₂ = 2.4 m/s²

The final velocity, v₂ = 0 m/s

The initial velocity, u₂ = v = 33 m/s

The time it takes to stop = t₂

-a₂ = 2.4 m/s²

∴ a₂ = -2.4 m/s²

From, v = u + a·t, we have;

v₂ = v + a₂·t₂

By plugging in the values of the variables, we have;

0 m/s = 33 m/s + (-2.4 m/s²) × t₂

∴ 2.4 m/s² × t₂ = 33 m/s

t₂ = 33 m/s/(2.4 m/s²) = 13.75 s

The time it takes to stop, t₂ = 13.75 seconds

7 0

3 years ago

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den301095 [7]

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2 years ago