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BabaBlast [244]

2 years ago

7

A pendulum Bob released from some initial height such that the speed of the bob at the bottom of the swing is 1.9m/s. What is th

e initial height of the bob?

1 answer:

Dmitriy789 [7]2 years ago

8 0

Answer:

h = 18.4 cm

Explanation:

Given that,

The speed of the bob at the bottom of the swing is 1.9m/s.

We need to find the initial height of the bob. Let it is h.

We can find it using the conservation of energy i.e.

$mgh=\dfrac{1}{2}mv^2$

Where

v is speed of the bob

So,

$h=\dfrac{v^2}{2g}\\\\h=\dfrac{(1.9)^2}{2\times 9.8}\\\\h= 0.184\ m$

or

h = 18.4 cm

So, the initial height of the bob is 18.4 cm.

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The brightest star in the night sky in the northern hemisphere is Sirius. Its distance from Earth is estimated to be 8.7 light y

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Answer: $5.11(10)^{13}miles$

Explanation:

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In other words: It is the distance that the light travels in a year.

This unit is equivalent to $5.879(10)^{12}miles$ , which mathematically is expressed as:

$1Ly=5.879(10)^{12}miles$

Doing the conversion:

$8,7Ly.\frac{5.879(10)^{12}miles}{1Ly}=5.11(10)^{13}miles$ This is the distance from Earth to Sirius in miles.

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At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr

svp [43]

The acceleration of the particle at time t is:
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The velocity of the particle at time t is given by the integral of the acceleration a(t):
$v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s$
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$x(t)=\int v(t) = \int (8t^3)=8 \frac{t^4}{4}=2t^4 ft$

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3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

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Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

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$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

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$a = (0.295)(0.2)$

$a = 0.059m/s^2$

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Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

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The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

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kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

<em>Now, form the mathematical expression of Kinetic Energy:</em>

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

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3 years ago