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1 year ago

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A 45-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistan

ce, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
_______m

1 answer:

Lubov Fominskaja [6]1 year ago

7 0

The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.

<h3>What is the height of the pole vaulter?</h3>

The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.

Potential energy = Change in kinetic energy

mgh = m(v - u)²/2

h = (v - u)²/2g

h = (10 - 1.1)²/2 * 9.8

h = 4.04 m.

In conclusion, the height is determined from the potential energy at that height.

Learn more about potential energy at: brainly.com/question/14427111

#SPJ1

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Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

3 years ago

Can someone help me???

LenaWriter [7]

Answer:

heliocentric is when the earth and the other planets revolves around the sun.

Explanation:

I'm not sure about which one but I do know what heliocentric is.

6 0

3 years ago

A pitcher throws a baseball that reaches the catcher in 0.75 s. The ball curves because it is spinning at an average angular vel

7nadin3 [17]

The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as

$\omega = \frac{\theta}{t}$

Here,

$\theta$ = Angular displacement

t = time

The angular velocity is given as

$\omega = 230rev/min$

PART A) The angular velocity in SI Units will be,

$\omega = 230rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega = \frac{23}{3}\pi rad/s \approx 24.08rad/s$

PART B) From our first equation we can rearrange to find the angular displacement then

$\theta = \omega t$

Replacing,

$\theta = (24.08)(0.75)$

$\theta = 18.06 rad$

4 0

3 years ago

A sled is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy isJ. (Formula: PE =

elixir [45]

Data:

h = 2m
m = 45 Kg
PE = ? (Joule)

Adopting, gravity (g) ≈ <span>9,8 m/s²
</span>
Formula: $PE_{grav} = mass * g * height$

Solving:

$PE_{grav} = mass * g * height$
$PE_{grav} = 45*9,8*2$
$PE_{grav} = 882J$

Answer:
<span>The sled's potential energy is 882 Joules</span>

8 0

3 years ago

Read 2 more answers

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago