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3 years ago

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A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

of the first at the moment the spring is extended to the end (x = A). Which of the following is true?
A. The maximum velocity does not change.
B. The amplitude of the oscillation A decreases.
C. The frequency of oscillations f does not change.
D. The frequency of oscillations f increases.
E. The period of oscillation increases.

1 answer:

elixir [45]3 years ago

8 0

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

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A car travels a distance of 400 m in 5 seconds. Calculate its average velocity.

ikadub [295]

Answer:

80 m/s

Explanation:

x = 400 m

t = 5 s

x = vt,

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3 years ago

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

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a machine gun fires 10 rounds per second the speed of the bullets is 300 m/s. what is the distance in the air between the flying

vovangra [49]

<span>(300 m/s)/(10 r/s) = 30 m/round.</span>

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nikklg [1K]

Answer:

The wavelength in miles is <u>0.1165 miles</u>.

Explanation:

Given:

Wavelength of the radio wave is 187.37 m.

Now, the wavelength is given in meters.

We need to convert the wavelength from meters to miles.

In order to convert meters to miles, we have to use their conversion factor.

We know that,

1 meter = $\frac{1}{1609}\ miles$

Therefore, the conversion factor is given as:

$CF=\frac{1}{1609}\ miles\ per\ meter$

So, the wavelength in miles is given as:

$Wavelength=\textrm{Wavelength in meters}\times CF\\\\Wavelength=187.37\ m\times \frac{\frac{1}{1609}\ miles}{1\ m}\\\\Wavelength=\frac{187.37}{1609}\ miles\\\\Wavelength=0.1165\ miles$

Hence, the wavelength in miles is 0.1165 miles.

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