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kondor19780726 [428]
3 years ago
12

A cartoon shows two friends watching an unoccupied car in free fall after it has rolled off a cliff. one friend says to the othe

r "It goes from zero to sixty miles per hour in about three seconds." is this statement correct?
Physics
2 answers:
Alona [7]3 years ago
6 0
The acceleration of gravity is 32.2 ft/sec^2.
After 3 seconds of free fall, the car is falling at 96.6 ft/sec.

60 mph is 88 ft/sec.

Let's look at it in two ways:

==> 96.6 is 9.8% greater than 88 .

==> It takes the car (88/32.2)=2.73 seconds to reach 60 mph. 3 is 9.8% greater than 2.73 .

Either way you look at it, the rough number is within 10% of the real number. That's plenty close enough to say that they're "about" the same.

It really depends on what the friends mean when they say "about". As an engineer, my personal feeling is that anything less than 10% off is a beautiful match.

enyata [817]3 years ago
3 0
1 mile = 1609 meters.60 miles = 96 540 meters.1 hour = 3 600 seconds.
Then:The velocity V = 96 540 / 3600 ≈ 26.8 m/s
Acceleration of free falla = V/t = 26.8/3 ≈ 8.9 m/c².
But it should be (excluding air resistance of 9.8 m/s².
Answer: the statement is wrong!




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An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

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We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

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a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

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3 0
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PLEASE HELP.. I WOULD BE SO HAPPY!! ILL MAKE YOLO THE BRAINLIEST....
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(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th
Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:  

V_{f}^{2} = V_{0}^{2} - 2gh

<u>Where:</u>

V_{f}: is the final speed = 14 m/s

V_{0}: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s  

b) The maximum height is:

V_{f}^{2} = V_{0}^{2} - 2gh

h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m

c) The time can be found using the following equation:

V_{f} = V_{0} - gt

t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s

d) The flight time is given by:

t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s

         

I hope it helps you!    

3 0
3 years ago
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