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Hatshy [7]
3 years ago
5

How long will a plane have to fly continuously with 1100 miles per hour in order to cover the same distance as that from Earth t

o the next star, Proxima Centauri?
Physics
1 answer:
vaieri [72.5K]3 years ago
4 0

Explanation :

It is given that,

Velocity of plane, v = 1100 mi/h

Proxima Centauri is a red dwarf star. Its distance from Earth to  Proxima Centauri is 4.243 light year.

Or we can say that the distance between Earth and  Proxima Centauri is,

d=2.49\times 10^{13}\ miles

We know that, time t=\dfrac{d}{s}

t=\dfrac{2.49\times 10^{13}\ miles}{1100\  miles/h}

t=0.22\times 10^{11}\ h

or

t = 2.5 years ( approx )

Hence, this is the required solution.

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which of the following types of exercise increase lean body mass?. A.) cardiorespiratory fitness. B.) flexibility. C.) muscular
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4 years ago
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The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental
pav-90 [236]

A) 5.0\cdot 10^{-11} m

The energy of an x-ray photon used for single dental x-rays is

E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J

The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m

4 0
3 years ago
Against the wind a commercial airline in south america flew 784 miles in 4 hours. with a tailwind the return trip took 3.53 hour
ira [324]

First let us assign variables,

d = distance travelled

t = time it took

v = velocity of the commercial airline

In linear physics, the equation for velocity is given as:

v = d / t

Rewriting for d:

d = v t

We know that the distance to and from south America are equal therefore:

d1 (going) = d2 (return)

Let us say that velocity of air is v3. Since going to South America, the wind is against the direction of the plane and the return trip is the opposite, therefore:

(v1 - v3) t1 = (v1 + v3) t2

(v1 – v3) 4 = (v1 + v3) 3.53

4 v1 – 4 v3 = 3.53 v1 + 3.53 v3

0.47 v1 = 7.53 v3

v1 = 16.02 v3

Since we also know that:

(v1 - v3) t1 = 784

(16.02 v3 – v3) * 4 = 784

60.085 v3 = 784

v3 = 13.05 mph

Therefore the speed of the plane in still air, v1 is:

v1 = 16.02 * 13.05

<span>v1 = 209.03 mph           (ANSWER)</span>

<span> </span>

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4 years ago
An apple is dropped from the top of a building and hits the ground 2 seconds later. How tall is the building?
Ket [755]

Hello! :)

\large\boxed{19.6 m}

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d = \frac{1}{2} at^{2}

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t = time (seconds)

Plug in the given values into the equation:

d = \frac{1}{2} (9.8)2^{2}\\\\d = \frac{1}{2} (39.2)\\\\d = 19.6 m

8 0
3 years ago
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