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2 years ago

6

A diver makes 0.50 revolutions on the way from a 9.6-m-high platform to the water. Assuming zero initial vertical velocity, find

the diver's average angular velocity during a dive.

1 answer:

devlian [24]2 years ago

7 0

Answer:

ω = 0.36 rev/s = 2.24 rad/s

Explanation:

First, we will find the time taken by the diver to reach the water. For this we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = height = 9.6 m

Vi = initial vertical velocity = 0 m/s

t = time taken = ?

g = 9.8 m/s²

Therefore,

9.6 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t = √1.95 s²

t = 1.4 s

Now, the average angular speed of diver will be:

ω = No. of Revolutions/t

ω = 0.5 rev/1.4 s

<u>ω = 0.36 rev/s = 2.24 rad/s</u>

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2 years ago

An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a s

Dmitriy789 [7]

Answer:

0.17724 m/s²

Explanation:

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2 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

6 0

2 years ago

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b. $v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

c.Work don=Final K.E-Initial K.E

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

4 0

2 years ago

Read 2 more answers

I need help on number 5. We need to use one of the four kinematics equations.

Gnom [1K]

ANSWER

$35.02m$

EXPLANATION

Parameters given:

Initial velocity, u = 26.2 m/s

When the vase reaches its maximum height, its velocity becomes 0 m/s. That is the final velocity.

We can now apply one of Newton's equations of motion to find the height:

$v^2=u^2-2as$

where a = g = acceleration due to gravity = 9.8 m/s²

Therefore, we have that:

$\begin{gathered} 0=26.2^2-2(9.8)s \\ \Rightarrow19.6s=686.44 \\ s=\frac{686.44}{19.6} \\ s=35.02m \end{gathered}$

That is the height that the vase will reach.

5 0

4 months ago