MULTIPLE CHOICE

How does the state of matter of the medium through which the sound waves and light waves travel theough affect their velocity

yanalaym [24]

Answer:

idk

Explanation:

idk

5 0

3 years ago

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

$\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}$

when the frequency is doubled

$\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1$

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

$f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}$

when the frequency is doubled;

$T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}$

Thus, the period will be reduced to one-half of what it was.

5 0

3 years ago

A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Natasha2012 [34]

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0

3 years ago

A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.0062

galina1969 [7]

Answer:

Force on the proton will be $.73\times 10^{-14}N$

Explanation:

We have given speed of proton is 20.3% of speed of light

Speed of light $c=3\times 10^8m/sec$

So speed of proton $v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec$

Magnetic field B = 0.00629 T

Charge on proton $q=1.6\times 10^{-16}C$

Angle between velocity and magnetic field $\Theta =137^{\circ}$

Force on the proton is equal to $F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N$

4 0

3 years ago

A negative point charge q1 = 25 nC is located on the y axis at y = 0 and a positive point charge q2 = 10 nC is located at y =14

sergey [27]

Answer:

y = 0.1 m

Explanation:

The electrical power for point loads is

V = $k \sum \frac{q_i}{r_i}$ k Sum qi / ri

in this case

V = k ( $- \frac{q_1}{r_1 } + \frac{q_2}{r_2}$ )

indicate that V = 0

$\frac{q_1}{r_1} = \frac{q_2}{r_2}$

r₂ = $\frac{q_2}{q_1} r_1$

the distance r1 is

r₁ = y -0

the distance r2

r₂ = 0.14 -y

we substitute

0.14 - y = $\frac{10}{25}$ y

y ( $\frac{10}{25} + 1$ ) = 0.14

y 1.4 = 0.14

y = 0.14 / 1.4

y = 0.1 m

7 0

3 years ago