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4 years ago

1. When an object's distance from another object is changing it must be

1 answer:

3 0

An object is in motion if its distance from another object is changing. An object is in motion if it changes position relative to a reference point. An reference point is a place or object used for comparison to determine if something is moving.

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A 90 kilogram diver jumped from a cliff. If it takes 2.26 seconds for the diver to hit the water, how high is the cliff?

Tasya [4]

The height of the cliff will be 25.05 m. The height of the cliff is obtained by the Newton equation of motion.

<h3>What is height?</h3>

The vertical distance between the object's top and bottom is defined as height. It is measured in centimeters, inches, meters, and other units.

The given data in the problem is;

Mass of diver(m)=90 kilogram

Time period(t)=2.26 sec

Initial velocity (u)=0 m/sec

The height attained by the cliff is;

$\rm H = ut + \frac{1}{2}gt^2 \\\\ H=\frac{1}{2}gt^2\\\\ H= \frac{1}{2}(9.81)(2.26)^2\\\\ H= \frac{1}{2} \times 9.81 \times 5.1076 \\\\ H=25.05 \ m$

Hence, the height of the cliff will be 25.05 m.

To learn more about the height, refer to the link;

brainly.com/question/10726356

#SPJ2

8 0

3 years ago

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Consider a hot air balloon rising vertically from a launch site located on the ground. A person is initially standing 10 m from

abruzzese [7]

Answer:

The distance between the person and the balloon after 2 seconds the person starts walking is changing on 4.47 m/s

Explanation:

The relative position between the balloon and the person is found using Galileo's relativity principle:

$\overrightarrow{r_{b/p}}=\overrightarrow{r_{b}}+\overrightarrow{r_{p}}$ (1)

with Rb the position of the balloon and Rp the position of the person respect with the origin (See Figure 1). Because we don’t have those positions but we know the constant velocities, we can relate the positions (R) with the velocities (v) with the kinematic equation:

$\overrightarrow{R}=\overrightarrow{v}*t$ (2)

So equation (1) is:

$\overrightarrow{R_{b/p}}=\overrightarrow{v_{b}*t}+\overrightarrow{R_{p}}=(v_{b}*t)\,\hat{j}-R_{p\,}\hat{i}$ (3)

with $\hat{j}$ and $\hat{i}$ the unitary vectors on y and x direction respectively.

We see from our initial condition (See figure 2) that:

$R_{p}=(10-v_{p}*t)$ (4)

So if we put this on (3) and divide by time we have:

$\frac{\overrightarrow{R_{b/p}}}{t}=\overrightarrow{v_{b/p}}=\frac{(v_{b}*\cancel{t})}{\cancel{t}}\,\hat{j}-\frac{(10-v_{p}*t)}{t}\hat{i}$ (5)

But we are interested in how fast a distance is changing, and that is a speed so:

$v_{b/p}=\sqrt{v_{b}^{2}+\left(\frac{(10-(v_{p}*t))}{t}\right)^{2}}=\sqrt{4^{2}+\left(\frac{(10-(3*2))}{2}\right)^{2}}\simeq4.47\,\frac{m}{s}$

3 0

3 years ago

A projectile is launched straight up at 135 m/s. Determine how fast it is moving at the top of its trajectory (flight).

suter [353]

<span> It is 0 m/s. At the very top of its flight is right between when it is going up and going down; this is the ephemeral moment when it is perfectly still. </span>

7 0

3 years ago

What causes wind to flow ?

Vladimir [108]

Answer:

Explanation:

as a result of the collision of hot and cold air

hope it helps

have a nice day

6 0

3 years ago

Read 2 more answers