Exercise provides a healthy outlet for feelings, which helps improve

Mazyrski [523]

Answer:3) variable affinities (stickiness) for something it is running past. Physical ... -measurement number (significant digits) unit (such as inches) -Significant ... Mass 1 oz. 28.25 g. Relations Between English and Metric Units Mass 1 dram. 1.772 g ... -graduated cylinder has an error of about 1% (± 0.1 mL in 10 mL). -Volumetric

Explanation:

3 0

3 years ago

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Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

3 years ago

A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=

VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

$U=\frac{1}{2}k (\Delta x)^2$

where

k is the spring constant

$\Delta x$ is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

$\Delta x = d-0=d$

and the amount of energy required is

$U=\frac{1}{2}k d^2$

In the second case, the spring is compressed from x=0 to x=-d, so

$\Delta x = -d -0 = -d$

and the amount of energy required is

$U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2$

so we see that the amount of energy required is the same.

7 0

3 years ago

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN

timurjin [86]

Answer:

$f_{o}$ = 391.67 Hz

Explanation:

The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency ( $f_{o}$ ).

The For a vibrating string, the fundamental frequency ( $f_{o}$ ) can be determined by:

$f_{o}$ = $\frac{v}{2L}$

Where v is the speed of waves of the string, and L is the length of the string.

L = 42.0 cm = 0.42 m

v = 329 m/s

$f_{o}$ = $\frac{329}{2*0.42}$

= $\frac{329}{0.84}$

$f_{o}$ = 391.6667 Hz

The fundamental frequency of the string is 391.67 Hz.

3 0

2 years ago

What are the different ways that the simulation shows you that the equation is balanced, visually? For each balanced reaction, i

Nastasia [14]

Answer:

e

Explanation:

e

3 0

3 years ago