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3 years ago

Mechanical energy is lost:

1 answer:

7 0

3. Due to the fact that friction is not converted to kinetic energy nor potential energy. The energy is converted into heat energy which is lost and can’t be put back

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Differentiate between angular displacement and linear displacement.

Sauron [17]

Answer:

The angular displacement is not a length (not measured in meters or feet), so an angular displacement is different than a linear displacement. ... As the object rotates through the angular displacement phi, the point on the edge of the disk moves distance sa along a circular path.

7 0

3 years ago

I need help on a it all please

Akimi4 [234]

your answer is the letter (b)

7 0

4 years ago

Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat

vichka [17]

The period of a pendulum is given by
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
$\frac{T_m}{T_e} = \frac{2 \pi \sqrt{ \frac{L}{g_m} } }{2 \pi \sqrt{ \frac{L}{g_e} } }= \sqrt{ \frac{g_e}{g_m} }$
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is $g_e = 9.81 m/s^2$ while on the Moon is $g_m=1.63 m/s^2$ , the ratio between the period on the Moon and on Earth is
$\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45$

3 0

3 years ago

What is the value of n in the balmer series for which the wavelength is 410.2 nm.?

m_a_m_a [10]

The answer is n= 6.

What is Balmer series?

The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.

To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.

To learn more about emission spectrum click on the link below:

brainly.com/question/24213957

#SPJ4

3 0

2 years ago

Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon

Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

$T^2=\frac{4\pi}{GM}r^3$

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

$T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min$

the comet takes around 16.63min

8 0

3 years ago