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3 years ago

A Person whose weight is 5.20 x 10^2 N is being pulledup

Physics

1 answer:

Gnoma [55]3 years ago

6 0

Answer:

Explanation:

Given

Weight of Person $W=5.20\times 10^{2} N$

Cave is $h=35.1 m$ deep

Breaking stress $T=569 N$

Net Force on Person

$F_{net}=569-520=49 N$

$a_{net}=\frac{F_{net}}{\frac{W}{g}}$

$a_{net}=\frac{49}{\frac{520}{9.8}}$

$a_{net}=0.923 m/s^2$

The shortest time such that the person can be taken out of cave

$h=ut+\frac{1}{2}at^2$

where

h=distance moved

t=time

a=acceleration

$35.1=0+\frac{1}{2}(0.923)(t)^2$

$t^2=76.05$

$t=\sqrt{76.05}$

$t=8.72\ s$

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The highest-pitched sound humans can hear have a wavelength of 0.017 m in air. What is the frequency of these sound waves if the

Verdich [7]

Answer: frequency is 20000hz

Explanation:

Velocity=340m/s

Wavelength=0.017m

Frequency=velocity ➗ wavelength

Frequency=340 ➗ 0.017

Frequency=20000hz

5 0

3 years ago

Read 2 more answers

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.12 m whose potential is 230

slamgirl [31]

Answer:

(a) q=3.07 nC

(b) σ=17 nC/m²

Explanation:

Given data

Radius r=0.12m

Potential V=230 V

To find

(a) Charge q

(b) Charge density σ

Solution

For Part (a)

As we know that potential is:

$V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}$

Substitute the given values

$q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC$

For Part (b)

The charge density is given by:

σ=q/(4πr²)

Substitute the given values and value of q to find charge density

$=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2$

σ=17 nC/m²

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3 years ago

What is the relationship between distance, time and acceleration

Evgesh-ka [11]

Answer:

total distance travelled

Explanation:

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4 years ago

When on object’s spectral lines are shifted from their rest wavelengths to longer wavelengths, we say that the object’s spectrum

neonofarm [45]

Answer:

Both statements are true.

Explanation:

When a celestial object (stars, galaxies) is moving away from an observer its spectral lines¹ will be shifted to the red part of the spectrum² (longer wavelength), in the other hand if the celestial body is moving toward the observer, the spectral lines will be shifted to the blue part of the spectrum (shorter wavelength). That is known as the Doppler shift.

This Doppler shift can be explained with the Doppler Effect³, which is defined for the case of light as:

$\frac{\Delta \lambda}{\lambda_{0}} = \frac{v}{c}$ (1)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the rest wavelength, v is the velocity of the source and c is the speed of light.

This redshift in distant galaxies is a strong evidence for the expansion of the universe. Equation 1 also allows the measurement of radial velocity from celestial objects.

Summary:

Blueshift:

$\lambda$

Redshift:

$\lambda >\lambda_{0}$

¹Spectral lines: Determines the presence of particular elements in the photosphere of an star.

²Spectrum: Decomposition of light in its characteristic colors (wavelengths).

³Doppler Effect: Change in the frequency of a wave as a consequence of the movement from a source relative to an observer or vice versa.

4 0

4 years ago

What best accounts for the periodic trends seen in ionization energy?

NeX [460]

Answer:

Atomic size

Explanation:

The Ionization energy relies on the atomic size, in light of the fact that with increment in size, the separation among core and valence electrons increments and subsequently attraction force among core and valence electron diminishes.

Ionization energy is the energy required to expel an electron from an impartial atom in its gaseous stage. The lower this vitality is, the more promptly the molecule turns into a cation. Consequently, the higher this energy is, the more impossible it is that the molecule turns into a cation.

By and large, components on the right side of the periodic table have a higher ionization vitality in light of the fact that their valence shell is almost filled.
Components on the left half of the periodic table have low ionization energies as a result of their readiness to lose electrons and become cations.
Another factor that influences ionization vitality is electron protecting. Electron protecting portrays the capacity of a molecule's internal electrons to shield its decidedly charged core from its valence electrons.
When moving to one side of a period, the quantity of electrons increments and the quality of protecting increments.
Accordingly, it is simpler for valence shell electrons to ionize, and in this manner the ionization vitality diminishes down a gathering. Electron protecting is otherwise called screening.
The ionization energy of the components inside a period for the most part increments from left to right. This is because of valence shell security.
The ionization energy of the components inside a gathering by and large diminishes through and through. This is because of electron protecting.
The noble gases have extremely high ionization energies in light of their full valence shells as demonstrated in the chart. Note that helium has the most noteworthy ionization vitality of the considerable number of components.

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3 years ago