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2 years ago

14

If an object starts at rest and moves 90 meters south along a straight line for 3 seconds, what is its average velocity?

2 answers:

ICE Princess25 [194]2 years ago

7 0

I believe the answer is B

MatroZZZ [7]2 years ago

7 0

Your answer is B. 30m/s in south.

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3 years ago

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

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$\Delta W = \frac{1}{2} mv^2$

Here,

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v = Velocity

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3 years ago

Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards Vernonville, and another car leaves Verno

iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is $v_{A} =\frac{x_{A} }{t_{A} }$ (Equation A) and Average speed of the car B is $v_{B} =\frac{x_{B} }{t_{B} }$ (Equation B), where $x_{A}$ and $x_{B}$ are the distances and $t_{A}$ and $t_{B}$ are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

$96 minutes . \frac{1 hour}{60 minutes} = 1.6 h$

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

$v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h}$ (Equation C)

$v_{B} = \frac{208-x}{1.6h}$ (Equation D)

From Equation A and C, we have:

$\frac{208-x}{1.6}+10 = \frac{x}{1.6}$

208-x+16 = x

208 + 16 = 2x

$x = \frac{224}{2}$

x = 112 miles

Replacing x in Equation A:

$v_{A} = \frac{112miles}{1.6h}$

$v_{A} = 70 miles per hour$

Replacing x in Equation B:

$v_{B} = \frac{208miles-112miles}{1.6h}$

$v_{B} = \frac{96miles}{1.6h}$

$v_{B} = 60 miles per hour$

3 0

3 years ago