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3 years ago

A cat is moving at 18 m/s when it accelerates at 4 m/s2 for 2 seconds. what is his new velocity?

Physics

2 answers:

larisa [96]3 years ago

8 0

Velocity =18
Velocity at 2 seconds=18+4+4=26

shutvik [7]3 years ago

6 0

Answer:

26m/s

Explanation:

Acceleration is the change in velocity of a body with respect to time. Mathematically,

Acceleration = change in velocity/time

Acceleration = final velocity (v) - initial velocity (u)/time (t)

Given a = 4m/s², v = ? u = 18m/s, t = 2seconds.

Substituting in the given formula,

4 = v-18/2

8 = v-18

v = 8+18

v = 26m/s

The new velocity of the body is therefore 26m/s

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dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

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vredina [299]

Answer:

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loris [4]

<u>Answer</u>

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<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;

Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
Ns ⇒number of turns in the seconndary coil
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Vs ⇒secondary voltage

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Vp = (120 × 4)/10

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