17. At what speed must an object be launched vertically upward for the
object to reach a maximum height of 16 meters above its starting position.
(Do not include units, do not include a sign, use 3 sig figs) *

Answers

Answer 1
Answer:   17.7 meters per second

===============================================

Explanation:

When starting at a height of 0, and initial velocity v, the equation of projectile motion is

y = -4.9x^2 + v*x

Where x is the time in seconds and y is the height at time x.

The vertex (h,k) of this parabola is the highest point. In general the vertex has its x coordinate at h = -b/(2a), so,

h = -b/(2a)

h = -v/(2*(-4.9))

h = v/(9.8)

Plug this into the first equation shown above. Also, plug in y = 16. Solve for v.

So we get the following.

y = -4.9x^2 + v*x

16 = -4.9(v/9.8)^2 + v*(v/9.8)

16 = (-5/98)v^2 + (5/49)v^2

16 = (-5/98)v^2 + (10/98)v^2

16 = (-5/98+10/98)v^2

16 = (5/98)v^2

(5/98)v^2 = 16

v^2 = 16*(98/5)

v^2 = 313.6

v = sqrt(313.6)

v = 17.708754896943

v = 17.7

The starting velocity must be roughly 17.7 meters per second.


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Answer:

Given that voltage is a measurement of energy per unit charge, Kirchhoff's loop rule is based on the law of conservation of energy, which states: the total energy gained per unit charge must equal the amount of energy lost per unit of charge.

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Answer:

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Answers

Answer:

I believe it is 600.

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Answer:

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Explanation:

Between the orbits of Mars and Jupiter, several thousand small objects called asteroids move in nearly circular orbits around the Sun. Consider an asteroid that is spherically shaped with radius r and density 2000 kg/m^3.


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Answers

Answer:

The radius is  [tex]r = 3.21*10^{4} \ m[/tex]

Explanation:

From the question we are told that  

   The density is  [tex]\rho = 2000 \ kg/m^3[/tex]

    The speed is  [tex]v = 24 \ m/s[/tex]

Generally the largest radius of the asteroid is mathematically represented as

      [tex]r = \frac{v^2}{ \sqrt{G * \rho * [\frac{4}{3} ] * \pi} }[/tex]

=>   [tex]r = \frac{24^2}{ \sqrt{ 6.67*10^{-11} * 2000 * [\frac{4}{3} ] *3.142 } }[/tex]

=>   [tex]r = 3.21*10^{4} \ m[/tex]

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Explanation:

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Answers

Answer:

0.231 rev/s

Explanation:

[tex]m[/tex] = Mass of cylinder

[tex]d[/tex] = Diameter of the cylinder = [tex]13.5\ \text{m}[/tex]

[tex]r[/tex] = Radius = [tex]\dfrac{13.5}{2}=6.75\ \text{m}[/tex]

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[tex]\mu N=mg\\\Rightarrow \mu\times mr\omega^2=mg\\\Rightarrow \omega=\sqrt{\dfrac{g}{\mu r}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.695\times 6.75}}\\\Rightarrow \omega=1.45\ \text{rad/s}[/tex]

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[tex]\dfrac{1.45}{2\pi}=0.231\ \text{rev/s}[/tex]

The number of revolutions of the cylinder is 0.231 rev/s.

PLEASE HELPPPP!!! I'VE BEEN STUCK ON THIS PROBLEM FOR A WHILE!!!!
A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km North Westtill she discovers the cave she wanted to see. Determine the magnitude and direction of her resultant displacement between the base camp and the cave.

Answers

Answer:

Resultant displacement is 29.2 km at north of west

Explanation:

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Taking east as positive x direction and north as positive y- direction, we have:

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- Second displacement is 41.0 km northwest, so its components are

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Answers

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Answer:

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Answers

Answer:

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Explanation:

Answer:

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Answers

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Answer:

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Explanation:

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Answers

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Answers

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