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yKpoI14uk [10]
3 years ago
9

if you take one hour drive at an average speed of 65 mph, is it possible for another car with an average speed of 55 mph to pass

you
Physics
2 answers:
SVETLANKA909090 [29]3 years ago
7 0
Nope, this is impossible because in order for a car to pass another, they the 55 mph car would have to be behind the 65 mph car (meaning that starting ahead of the 65 mph doesn't count as pass). Assuming that they both drive for one hour, it is impossible because the first will cover a distance of 65 mi and the second would cover a distance of 55 mi. One is obviously ahead of the other and is therefore impossible for the slow one to pass the first one unless the slow car keeps driving after an hour. In that case, it would take approximately 11 minutes for it to pass the other car. This was found by finding the distance needed to pass the first car : 65 - 55 = 10 mi and converted using 1 hr/ 55 mi = .18181818 hr x 60 min/ 1 hr = 11 seconds
I hope this helps :)
Aleks04 [339]3 years ago
3 0
Sure it is.  We only know the average speeds, but we don't know
how much of the time each car drove faster or slower. 

The averages only tell us who got to the end of their trip sooner. 
But there could have been a time ... or several times during the hour ...
when the 55mph car was blazing and passed the 65mph car.

In fact, the 55mph car could have been driving at 110 mph for the
first 29 minutes, then the last 1.83 miles at a speed of 3.5 mph.
His average would then be 55mph, even though he spent the first
29 minutes passing everybody and everything on the road in the
first 53.16 miles of his trip.
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Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

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You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
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The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
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Let's use the other equation to find the time
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              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

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4 years ago
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If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

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<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0
3 years ago
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