1. Approximately how many people watch the March Madness tournament?

The graph below represents the relationship between the force exerted on an elevator and the distance the elevator is lifted.

Stella [2.4K]

I have three problems with this question.

#1). If you copied the question exactly the way it's written,
then the question is written very badly. The wording is
misleading, and the more you try to think about it and
puzzle it out, the more it'll damage your understanding
of Physics.

There is no relationship between the force exerted on an
elevator and the distance the elevator is lifted.

-- If the force is anything more than the weight of the elevator ...
even one ounce more ... then it'll lift the elevator as high as
you want.

-- If the force is anything less than the weight of the elevator ...
even one ounce less, then that elevator is headed for the bottom.

#2). You didn't post any graph below, so if we need the graph
to answer the question, then we can't answer the question.

#3). I guess that's OK, because you didn't ask any question.

8 0

3 years ago

Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano

Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

Specific energy is given by

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$

Here

ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2$

Value of specific energy is also given as

$\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km$

Orbit formula is given as

$r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0$

Putting values in this equation and solving for e via the quadratic formula gives

$ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69$

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

The radius of trajectory at perigee is given as

$r_p=a(e-1)\\$

Substituting values gives

$r_p=133319 (1.086-1)\\r_p=11465.4 km$

Now for estimation of altitude z above earth is given as

$z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km$

So the altitude at closest approach is 5088 km

radius of perigee is also given as

$r_p=\frac{h^2}{\mu}\frac{1}{1+e}$

Rearranging this equation gives

$h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s$

Now the velocity at perigee is given as

$v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\$

So the velocity at perigee is 8.516 km/s

Turn angle is given as

$\delta =2 sin^{-1} (\frac{1}{e})$

Substituting value in the equation gives

$\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08$

Aiming radius is given as

$\Delta =a \sqrt{e^2-1}$

Substituting value in the equation gives

$\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km$

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0

4 years ago

An aircraft is at a standstill. It accelerates to 140kts in 28 seconds. The aircraft weighs 28,000 lbs. How many feet of runway

stiv31 [10]

Answer:

runway use is 3307.8 feet

Explanation:

given data

velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s

time = 28 seconds

weight = 28000 lbs

to find out

How many feet of runway was used

solution

we will use here first equation of motion for find acceleration

v = u + at ..............1

here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time

put here value in equation 1

72.016 = 0 + a(28)

a = 2.572 m/s²

and

now apply third equation of motion

s = ut + 0.5×a×t² .......................2

here s is distance and u is initial speed and t is time and a is acceleration

put here all value in equation 2

s = 0 + 0.5×2.572×28²

s = 1008.24 m = 3307.8 ft

so runway use is 3307.8 feet

5 0

4 years ago

A gas is enclosed in a confainer fitted with a piston of cross sectional area 0.10 the pressureof the gas is maintained in 8000

Arlecino [84]

Answer:

10 Joule

Explanation:

The solution and answer are well written in the Pic above.

6 0

3 years ago

A scientist heats an unknown substance in a closed system. The graph shows the change in temperature over time.

jeka94

Answer:

A

Explanation:

It's A

8 0

3 years ago

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