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3 years ago

6

find the rate of positive acceleration of an automobile which went from a complete stop to a velocity of 30 meters per second in

10 seconds

1 answer:

maks197457 [2]3 years ago

3 0

Answer:

3 m/s^2

Explanation:

acceleration= Change in velocity/time

= 30-0 / 10

= 30/10

=3 m/s^2

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Answer:

v_g,i = 1.208 m/s

Explanation:

We are given;

Mass of girl; m_g = 47.2 kg

Mass of plank; m_p = 177 kg

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Let the velocity of plank to ice be v_p,i

Since the velocity of the girl is 1.53 m/s relative to the plank, then;

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From conservation of momentum;

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Thus;

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Dividing both sides by 47.2 gives;

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v_g,i + ((v_g,i)/3.75) = 1.53

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Read 2 more answers

A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?

iris [78.8K]

Answer:

$\Delta x=22.67786838m$

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

$y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2$ (1)

Where:

$y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$v_o=initial\hspace{3}velocity$

$t=travel\hspace{3}time$

$g=gravity\hspace{3}constant$

$\theta=Initial\hspace{3}launch\hspace{3}angle$

In this case:

$\theta=0$

Because the dog jumps horizontally

Let's asume the gravity constant as:

$g=9.8$

$y=0$

Because when the dog reach the base the height is 0

$y_o=70$

$v_o=6$

Now let's replace the data in (1)

$y_o-\frac{1}{2} *(9.8)*t^2+70$

Isolating t:

$t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473$

Finally let's find the horizontal displacement using this equation:

$\Delta x=v_o*cos(\theta)*t$

Replacing the data:

$\Delta x=6*1*3.77964473=22.67786838m$

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2 years ago