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3 years ago

Which of the following is closest to 2cm?

Physics

1 answer:

Sloan [31]3 years ago

4 0

C because the the smallest thing then the other ones because it never said what kind of size of it

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A wave that is traveling fast can be said to have a high ___

S_A_V [24]

A wave that is traveling fast can be said to have a high speed.<em> (b) </em>

Just like a car, motorcycle, or freight train that is traveling fast.

3 0

4 years ago

Does coefficient of thermal expansion vary with temperature?

N76 [4]

Answer:

yes

Explanation:

8 0

3 years ago

How much work is done if you push a box 200 meters with a force of 35 newtons answer?

VashaNatasha [74]

Work = force × distance
= 35 N × 200 m
= 7000 J

4 0

3 years ago

The focal length of the lens of a simple digital camera is 40 mm, and it is originally focused on a person 25 m away. In what di

ss7ja [257]

Answer:

Explanation:

Here image distance is fixed .

In the first case if v be image distance

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

In the second case

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )

3 0

4 years ago

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