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2 years ago

The moon is 250,000 miles away. How many feet is it from earth

Physics

2 answers:

beks73 [17]2 years ago

4 0

(250,000 miles) x (5,280 feet/mile) = 1,320,000,000 feet

( 1.32 x 10⁹ )

Fynjy0 [20]2 years ago

3 0

The conversion between miles and feet is:

$1 mi = 5280 ft$

So we can set up a proportion to find how many feet corresponds to 250000 miles:

$1 mi: 5280 ft = 250000 mi :x$

and solving it, we find

$x=\frac{5280 ft \cdot 250000 mi}{1 mi} =1.32 \cdot 10^9 ft$

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In a particular beam of radiation, which is traveling in a vacuum, the amounts of energy per second at an ultraviolet wavelength

Darina [25.2K]

Answer: d)

Explanation: In order to justify the answer we have to consider that the energy of photons directely depent on the frequency so the energy is inverselly dependent of the wavelegth.

If both beams have the same power, this means Energy/time so the number of photons per second must be different. As consequence a) is wrong as b) since it is not posible since UV photon have more energy that IR photons. c) It is no necessary know the frequency since the wavelength is related in the form:

c=λν c is the speed of light, λ the wavelegth and ν the frequency.

d) Certainly will be more more IR photons than UV photons to get the same beam power.

8 0

2 years ago

If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how

Arturiano [62]

Answer:

Vx= 11.0865(m/s)

Vy= 6.4008(m/s)

Explanation:

Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:

V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)

The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:

Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)

And for the same theorem the speed on the Y axis will be:

Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)

5 0

3 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

2 years ago

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du

Lana71 [14]

Answer:

$a_E=\dfrac{Gm}{r^2}$

Explanation:

$M_E$ = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

$M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}$

The acceleration that the Earth will feel is $a_E=\dfrac{Gm}{r^2}$

7 0

2 years ago

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.0 m above the water. One diver runs of

Alika [10]

Answer:

Number of revolutions=1.532 revolutions

Explanation:

Given data

Distance s=8.0 m

Angular speed a=1.2 rev/s

To find

Number of revolutions

Solution

From the equation of simple motion we not that

$S=ut+1/2gt^{2}\\ where\\u=0\\So\\8.0m=0+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{8.0m}{0.5*9.8m/s^{2} } \\ t^{2}=1.63\\t=\sqrt{1.63} \\t=1.28s$

So for the number of revolutions she makes is given as

$n=a*t\\n=(1.2rev/s)(1.28s)\\n=1.532revolutions$

8 0

3 years ago