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3 years ago

How much time does it take for a plane To fly 5000 miles if the plane travels a speed of 500 miles per hour?

Physics

2 answers:

Tasya [4]3 years ago

6 0

10 hours 500 x 2 = 1000 ...so every 2 hours = 1000 x 5 = 5000

Vitek1552 [10]3 years ago

5 0

Answer:

10 hours

Explanation:

If the plane flies at 500 miles per hour, it flies 500 miles every hour. So, to find out how many hours it will take to fly 5000, you simply do 5000/500.

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After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume t

Ivenika [448]

Time taken by the package to reach the sea level= 13.7 s

height=h=925 m

initial velocity along vertical= vi=0

acceleration due to gravity=g=9.8 m/s^2

using the kinematic equation h= Vi*t + 1/2 gt^2

925=0(t)+1/2 (9.8)t^2

4.9 t^2=925

t= 13.7 s

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3 years ago

WHAT SHOULD I NAME MY DEAD RAT?

olga_2 [115]

Answer:

pablito

Explanation:

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2 years ago

a coin is dropped from the top of a tall building. determine the coin's (a) velocity and (b) displacement after 1.5 sec.

Lilit [14]

1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters

8 0

2 years ago

The figure in Figure 1 shows two single-slit diffraction patterns. The distance between the slit and the viewing screen is the s

V125BC [204]

Answer:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."

Explanation:

The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.

D sin θ = m λ

if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."

This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.

Hope this helps!

8 0

1 year ago

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

3 years ago