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3 years ago

Make the following conversion. 24.7 mL = _____ cm3 247 2470 2.47 24.7

Physics

2 answers:

AnnyKZ [126]3 years ago

5 0

Since 1ml = 1 cm cubed its gonna be 24.7

Anna007 [38]3 years ago

3 0

24.7.....................................

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block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s). What is the angular frequen

faltersainse [42]

Answer:

Angular frequency is 20 rad/s.

Explanation:

Given that,

A block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s) is given by :

$x(t)=1.5\cos(20\ t)$ .....(1)

The general equation of oscillating particle is given by :

$x(t)=A\cos(\omega t)$ .......(2)

Compare equation (1) and (2) we get :

$\omega=20\ rad/s$

So, the angular frequency of the oscillation is 20 rad/s.

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3 years ago

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Explanation:

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4 years ago

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A toy train rolls around a horizontal 1.0-m-diameter track. the coefficient of rolling friction is 0.10. how long does it take t

Musya8 [376]

Deceleration due to rolling resistance (α) = coefficient*g/R = 0.1*9.81/0.5 = 1.962 rad/sec^2

Equations of motion:
ωf=ωo+αt
But ωf = 0 rad/sec; ωo= 2*pi*RPM/60 = 4.19 rad/sec

Therefore,
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3 years ago

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An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) what is its angular acceleration in rad/s2 ? (b) what i

MissTica

(a) The angular acceleration is given by:

$\alpha=\frac{\omega_f-\omega_i}{\Delta t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

$\Delta t$ is the time interval

Let's convert the angular velocities in rad/s and the time in seconds:

$\omega_f=100000 \frac{rev}{min}=100000 \frac{rev}{min} \frac{2 \pi rad/rev}{60 s/min}=10467 rad/s$

$\omega_i=0$

$\Delta t=2.00 min \cdot 60 \frac{s}{min} = 120 s$

Substituting the numbers into the equation, we find the angular acceleration:

$\alpha=\frac{10467 rad/s}{120 s}=87.2 rad/s^2$

(b) The tangential acceleration is given by the product between the angular acceleration and the distance of the point from the axis of rotation. In this case, the distance is

$r=9.5 cm=0.095 m$

Therefore the tangential acceleration is

$a=\alpha r=(87.2 rad/s^2)(0.095 m)=8.3 m/s^2$

(c) The radial acceleration (also known as centripetal acceleration) is given by the product between the square of the angular velocity and the distance from the axis of rotation:

$a=\omega^2 r=(10467 rad/s)^2(0.095 m)=1.04 \cdot 10^7 m/s^2$

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4 years ago