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2 years ago

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Sofia goes on a hike on a trail that is 10 km long. She starts at 2:00pm and ends at 5:00pm. The end of the trail is 300m north

of the beginning of the trail. What is Sofia's average velocity? (Velocity= displacement/time) Question 12 options: 100 m/hr S 30 m/hr S. 01 m/hr S 3. 3 m/hr S.

1 answer:

vlabodo [156]2 years ago

4 0

As the question depicts the trail is 10Km long but the end of the trail is 300m north from the beginning, Here the average velocity comes out to be 100m/hr.

<h3>What is Velocity?</h3>

Velocity is defined as the displacement covered by an object or a body in unit time interval.

From the problem, it is clear that the distance is 10Km but the displacement comes out to be 300m.

The time from 2:00 pm to 5:00 pm is a total of 3hours.

Now, the velocity is determined from the following formula

$V=\dfrac{d}{t}$

here,

d= displacement

t= time

So, by putting values in above formula we get,

$V=\dfrac{300 m}{3 hrs}$

$V=100m/hr$

So, the average velocity of Sofia is 100m/hr

To Know more about velocity

brainly.com/question/862972

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3 years ago

Read 2 more answers

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3 years ago

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2 years ago

Imagine that asteroid A that has an escape velocity of 50 m/s. If asteroid B has twice the mass and twice the radius, it would h

padilas [110]

Answer:

The same as the escape velocity of asteorid A (50m/s)

Explanation:

The escape velocity is described as follows:

$v=\sqrt{\frac{2GM}{R}}$

where $G$ is the universal gravitational constant, $M$ is the mass of the asteroid and $R$ is the radius

and since the scape velocity is 50m/s:

$50m/s=\sqrt{\frac{2GM}{R}}$

Now, if the astroid B has twice mass and twice the radius, we have that tha mass is: $2M$

and the radius is: $2R$

inserting these values into the formula for escape velocity:

$v=\sqrt{\frac{2G(2M)}{2R} } =\sqrt{\frac{4GM}{2R} } =\sqrt{\frac{2GM}{R} }$

and we have found that $50m/s=\sqrt{\frac{2GM}{R}}$ , so the two asteroids have the same escape velocity.

We found that the expression for escape velocity remains the same as for asteroid A, this because both quantities (radius and mass) doubled, so it does not affect the equation.

The answer is

Asteroid B would have an escape velocity the same as the escape velocity of asteroid A

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3 years ago