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3 years ago

A wave with a greater amplitude will transfer . . . . \

Physics

1 answer:

BabaBlast [244]3 years ago

6 0

less mass is more mass but less energy in more mass. less mass has more energy

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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago

I'd like you to think back on 2 events in

strojnjashka [21]

Answer:

Holi

biwali

these are best events and love to celebrate withmy family and friends it contains lot of happiness and joy

7 0

2 years ago

Describe how water may be treated before people use it

brilliants [131]

Water is treated by purifying it by adding slaked lime or potash alum

3 0

3 years ago

Read 2 more answers

A train starts at v=10m/s and accelerates at 1m/s2 for 5s. Its final speed is *

ad-work [718]

Answer:

D) 15 m/s

Explanation:

v = u + at

v = 10 m/s + 1 m/s²(5 s)

v = 15 m/s

3 0

2 years ago

The FM radio band in most places goes from frequencies of about 89 MHz to 106 MHz. How long is the wavelength of the radiation a

ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

8 0

3 years ago