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3 years ago

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A majorette in the Rose Bowl Parade tosses a baton into the air with an initial angular velocity of 2.5 rev/s. If the baton unde

rgoes a constant acceleration while airborne of 0.2 rev/s2 and its angular velocity is 0.8 rev/s when the majorette catches it, how many revolutions does it make in the air?

1 answer:

oksian1 [2.3K]3 years ago

4 0

Answer:

14 rev

Explanation:

$w_{o}$ = initial angular velocity = 2.5 revs⁻¹

$w$ = final angular velocity = 0.8 revs⁻¹

$\alpha$ = Angular acceleration = - 0.2 revs⁻²

$\theta$ = Angular displacement

Using the equation

$w^{2} = w_{o}^{2} + 2 \alpha \theta\\0.8^{2} = 2.5^{2} + 2 (- 0.2) \theta\\ \theta = 14 rev$

So the number of revolutions are 14

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Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col

Ratling [72]

In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
$p_f=(m+M)v_f=4.8 kg m/s$
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find $v_f$ , the velocity of the nail and the block after the collision:
$v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}= 0.48 m/s$

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
$p_i = mv_i = p_f$
Rearranging the formula, we can find $v_i$ , the velocity of the nail before the collision:
$v_i = \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s$

6 0

3 years ago

Read 2 more answers

How many sig figs are in 3252.6

jekas [21]

The number 3252.6 has 5 significant figures

7 0

3 years ago

A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo

gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

Height at any time 'T' after the drop =

(initial height) +
      (initial velocity) x (T) +
   (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

   Upward = the positive direction

   Initial height = +150 m
   Initial velocity = + 3 m/s

   Downward = the negative direction

   Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

   -4.9 T² + 3T + 150 = 0

Use the quadratic equation:

   T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]

   = (-1/9.8) [ -3 plus or minus 54.305 ]

   = (-1/9.8) [ 51.305 or -57.305 ]

      T = -5.235 seconds or 5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   H'(t) = v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set    v₀ + a T = 0

+3 - 9.8 T = 0

Add 9.8 to each side: 3   = 9.8 T

Divide each side by 9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh ! I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
   (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .

Distance climbed during that time = (average speed) x (time)

   = (1.5 m/s) x (0.306 sec)

   = 0.459 meter (hardly any at all)

   But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0

3 years ago

The number of significant figures in 0.060900 is

Lady bird [3.3K]

Answer:

There are <u>5</u> significant figures.

Explanation:

You must start conting your sig figs until you continue to hit zeros at the end. Those zeroes at the end are disregarded. So 0.0609 is where you get your <em>sig figs</em> from.

8 0

3 years ago

Azurite is a mineral that contains 55.1% of copper. How many meter of copper wire with diameter of 0.0113 in can be produced fro

soldier1979 [14.2K]

Answer:

1402.73 m

Explanation:

Mass of Azurite=3.25 lb

Percent of copper in AZurite mineral=55.1%

Diameter of copper wire,d=0.0113 in

Radius of copper wire= $r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in$

$\frac{1}{1000}=10^{-3}$

Density of copper= $\rho=8.96g/cm^3$

1 lb=454 g

3.25 lb= $3.25\times 454=1475.5 g$

Mass of Azurite= $1475.5 g$

Mass of copper= $\frac{55.1}{100}\times 1475.5=813 g$

Density= $\frac{Mass}{volume}$

Using the formula

$8.96=\frac{813}{volume\;of\;copper}$

Volume of copper wire= $\frac{813}{8.96}=90.7cm^3$

Radius of copper wire= $5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm$

1 in=2.54 cm

Volume of copper wire= $\pi r^2 h$

$\pi=3.14$

Using the formula

$90.7=3.14\times (14.35\times 10^{-3})^2\times h$

$h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}$

$h=140273 cm$

1 m=100 cm

$h=\frac{140273}{100}=1402.73 m$

Hence, the length of copper wire required=1402.73 m

7 0

3 years ago