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adoni [48]
3 years ago
6

You drive to the city at an average speed of 40 km/h and return at an average speed of 60 km/h. Find your average speed for the

entire trip? Explain why the answer is not 50 km/h.
Physics
1 answer:
Eva8 [605]3 years ago
7 0
Well, let's see:

If you live 'D' kilometers from the city, and you go and return by the same route, then

-- The time it takes you to reach the city, at 40 km/h, is  D/40  hours.

-- The time it takes you to return, at 60 km/h, is    D/60.

-- The total distance you cover is  2D, and the total driving time is

                        (D/40) + (D/60)

                   =  (3D/120)  +  (2D/120)

                   =        5D/120  =  D/24

-- Your average speed for the entire trip is

                         (distance covered)  /  (time to cover the distance)

                   =            ( 2D )   /   ( D/24 )

                   =              48 km/hr.  

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3 years ago
A place kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear
trapecia [35]
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4 0
4 years ago
A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl
nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

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7 0
3 years ago
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

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\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

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\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

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Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

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6 0
3 years ago
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