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3 years ago

5

You are riding your bike to the mall. You travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an ex

ample of
average velocity.

positive acceleration.

instantaneous speed.

negative acceleration.

2 answers:

vredina [299]3 years ago

5 0

Answer:

This is an example of negative acceleration.

Explanation:

It is given that,

You are riding your bike to the mall. You travel the first mile in 10 minutes. The last mile takes you 15 minutes.

The rate of change of velocity of an object is called its acceleration. In this case, the person is slowing down his bike. This shows that the person is decelerating i.e. it is an example of negative acceleration.

Hence, the correct option is (d) " negative acceleration ".

ratelena [41]3 years ago

4 0

I am pretty sure that when you travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an example of negative acceleration. I consider this to be correct because <span>the second mile was slower. Hope you will agree with me. Regards!</span>

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4. Determine the net force for the free-body

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3 years ago

A three-wheeled car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed

DiKsa [7]

Answer:

(a) Total time = 105.5 sec

(b)average speed = 31.26 m/sec

Explanation:

We have given that three-wheeled car starts from rest that initial velocity u = 0 m/sec

Final velocity = 35 m/sec

Acceleration $a=2m/sec^2$

According to first law of motion we know that v = u+at

So $35=0+2\times t$

t = 17.5 sec

After this car travels 83 sec at a constant speed and after that it takes 5 sec additional time to stop

(a) So total time in which car is in motion = 17.5+83+5 = 105.5 sec

(b) Total distance traveled during first 17.5 sec

$s=ut+\frac{1}{2}at^2=0\times 17.5+\frac{1}{2}\times 2\times 17.5^2=306.25m$

Distance traveled in 83 sec with with velocity of 35 m/sec = $35\times 83=2905m$

For next 5 second

Initial velocity u = 35 m/sec

Final velocity = 0 as finally car stops

So $a=\frac{v-u}{t}=\frac{0-35}{5}=-7m/sec^2$

Distance traveled $s=ut+\frac{1}{2}at^2=35\times 5+\frac{1}{2}\times -7\times 5^2=87.5m$

So total distance traveled = 306.25+2905+87.5 = 3298.75 m

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3 years ago

An applied force of 20 N is used to accelerate an object to the right across a

vichka [17]

Answer:

$F_{norm} = 100 N$

$F_{net}=10 N$

$\mu = 0.10$

$m = 10 kg$

$a=1.0 m/s^2$

Explanation:

To determine the normal force, we just need to analyze the situation along the vertical direction.

The box along the vertical direction is in equilibrium, so the equation of the forces is

$F_{norm} - F_{grav} = 0$

which means that

$F_{norm} = F_{grav} = 100 N$

The net force can be determined by looking at the situation along the horizontal direction (since the net force in the vertical direction) is zero. Here we have:

- An applied force of 20 N forward, $F_{app} = 20 N$

- A frictional force of 10 N backward, $F_{frict} = 10 N$

So, the net force is

$F_{net}=F_{app}-F_{frict}=20-10 = +10 N$ in the forward direction

The expression for the frictional force is

$F_{frict} = \mu F_{norm}$

where $\mu$ is the coefficient of friction. Solving for $\mu$ ,

$\mu = \frac{F_{frict}}{F_{norm}}=\frac{10}{100}=0.10$

The force of gravity is given by

$F_{grav}=mg$

where m is the mass of the object and $g=10 m/s^2$ . Solving for m, we find the mass of the object:

$m=\frac{F_{grav}}{g}=\frac{100}{10}=10 kg$

Finally, the acceleration can be found by using Newton's second law

$F_{net} = ma$

where a is the acceleration. Solving for a,

$a=\frac{F_{net}}{m}=\frac{10}{10}=1.0 m/s^2$

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3 years ago

If a person traveled from Sacramento to Los Angeles by car, their distance would be 384.9 miles. What would their displacement b

kumpel [21]

Answer:

Displacement from Sacramento to Los Angeles is 384.9 miles

Total path length is given as 769.8 miles

Displacement of complete round trip is ZERO

Explanation:

As we know that the displacement is the change in the position from initial to final in a straight line

Here we know that the car travel from Sacramento to Los Angeles

So the straight line distance between two is the displacement

so we have

displacement = 384.9 miles

Now we know that he returns from Los Angeles to Sacramento by same path

So total distance moved by it is same as total path length

d = 384.9 + 384.9

d = 769.8 miles

Also we know that initial and final positions are same so displacement is ZERO for complete round trip.

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