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3 years ago

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A car is driving down the highway at a 50 m/s. Suddenly, the driver sees an accident and slams on his breaks, coming to a comple

te stop in 10 seconds. How far did the car travel before coming to a stop?

1 answer:

vlada-n [284]3 years ago

4 0

Given parameters:

Speed of the car = 50m/s

Time = 10s

Unknown:

Distance traveled before the car stopped = ?

We need to establish speed - time - distance relationship.

Speed is the rate of change of distance with time. It is a scalar quantity .

Mathematically;

Speed = $\frac{distance}{time}$

Distance = speed x time

Input the parameters and solve;

Distance = 50 x 10 = 500m

The car traveled 500m before it stopped.

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A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 59 ms. (a

xxTIMURxx [149]

Answer: $y_{com}=0.707 m$

$v_{com}=3.713 m/s$

Explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by first stone in 430 ms

$y_1=ut+\frac{at^2}{2}$

$y_1=0+\frac{g(0.43)^2}{2}$

$y_1=0.9069 m$

Distance traveled by stone 2 in t=430-59=371 ms

$y_2=ut+\frac{at^2}{2}$

$y_2=0+\frac{g(0.0.371)^2}{2}$

$y_2=0.674 m$

velocity of first stone after t=0.43 s

$v_1=u+at$

$v_1=0+9.8\times 0.43=4.214 m/s$

velocity of second stone after t=0.371 s

$v_2=u+at$

$v_2=0+9.8\times 0.371=3.63 m/s$

Position of Center of mass of system

$y=\frac{y_1m_1+y_2m_2}{m_1+m_2}$

$y=\frac{0.9069\times m+0.674\times 6m}{m+6m}$

$y=\frac{4.95m}{7m}=0.707 m$

Velocity of COM

$v_{com}=\frac{v_1m_1+v_2m_2}{m_1+m_2}$

$v_{com}=\frac{4.214\times m+3.63\times 6m}{m+6m}$

$v_{com}=3.713 m/s$

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What is a tool that can be used to measure the size of a force?

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3 years ago

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Answer: c a d b

Explanation:

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3 years ago

A child bounces a 51 g superball on the sidewalk. The velocity change of the super bowl is from 22 m/s downward to 14 m/s upward

noname [10]

Answer:

$F=1.02x10^{-3} N$

Explanation:

From the exercise we know:

$m=51g*\frac{1kg}{1000g}=0.051kg$

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$v_{2}=14m/s$

$t_{2}-t_{1}=1800s$

So, the average acceleration is:

$a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{(14-(-22))m/s}{1800s}=0.02m/s^2$

The average force is:

$F=m*a=(0.051kg)(0.02m/s^2)=1.02x10^{-3} N$

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Answer:

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$v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s$

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This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

$s_h = v_ht = 1.8*0.293 = 0.528 m$

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3 years ago