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2 years ago

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A 15kg hoop with a radius of 3 m is rolling at on a level surface when it reaches a 25 degree incline. If it reaches a height of

4 m, what was the hoop's initial velocity?

1 answer:

dimulka [17.4K]2 years ago

4 0

The initial velocity of the hoop is determined as 8.854 m/s.

<h3>Conservation of energy</h3>

The initial velocity of the hoop can be determined from the principle of conservation of energy.

Final potential energy = Initial kinetic energy

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

2gh = v²

√2gh = v

√(2 x 9.8 x 4) = v

8.854 m/s = v

Thus, the initial velocity of the hoop is determined as 8.854 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

#SPJ1

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A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

3 years ago

What is your volume of the object?<br> 30 cm3<br> 35 cm3<br> 42 cm3<br> 54 cm3

Alinara [238K]

The answer is 54 cm3

3 0

3 years ago

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

krok68 [10]

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

3 0

3 years ago

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity a

Marina86 [1]

Answer:

$I_{2}=2.39*10^{-5} W/m^{2}$

Explanation:

The definition of the intensity in terms of power is given by:

$I=\frac{P}{A}$

Where:

P is the power
A is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is $A = 4\pi R^{2}$

To the first location we have:

$I_{1}=3*10^{-4} W/m^{2}=\frac{P}{4\pi 22^{2}}$

and to the second location we have:

$I_{2}=\frac{P}{4\pi 78^{2}}$

Now, we can divide each intensity to find the second intensity.

$\frac{I_{2}}{I_{1}}=\frac{\frac{P}{4\pi 78^{2}}}{\frac{P}{4\pi 22^{2}}}$

$I_{2}=I_{1}* \frac{22^{2}}{78^{2}}$

$I_{2}=3*10^{-4}\frac{22^{2}}{78^{2}}$

$I_{2}=2.39*10^{-5} W/m^{2}$

I hope it helps you!

5 0

3 years ago

Read 2 more answers

Anaerobic metabolism:

goblinko [34]

Explanation:

B. leads to muscle strain.

5 0

3 years ago