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2 years ago

7

the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds

s

1 answer:

ki77a [65]2 years ago

4 0

Answer:

Explanation:

The question is incomplete.

The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.

s= sin2(pi)t

Acceleration = d²S/dt²

dS/dt = 2πcos2πt

d²S/dt² = -4π²sin2πt

A(t) = -4π²sin2πt

Next is to find acceleration after 4.5 seconds

A(4.5) = -4π²sin2π(4.5)

A(4.5) = -4π²sin9π

A(4.5) = -4π²sin1620

A(4.5) = -4π²(0)

A(4.5) = 0m/s²

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2 years ago

Consider a series LRC-circuit in which C-120.0 uF. When driven at a frequency w = 200.0 rad s-1 the com ples impedance is given

Lina20 [59]

Answer:

(a). (i). The reactants are $X_{L} =31.66\ \Omega$ .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.

Explanation:

Given that,

Capacitor = 120.0 μC

Frequency = 200.0 rad/s

Impedance = 100.0 -10j

(I). We need to calculate the $X_{C}$

$X_{C}=\dfrac{1}{C\times\omega}$

Put the value into the formula

$X_{C}=\dfrac{1}{120\times10^{-6}\times200}$

$X_{C}=41.66\ \Omega$

(II). We know that,

Formula of impedance is

$Z=\sqrt{R^2+X_{L}^2+X_{C}^2}$ ...(I)

Given equation of impedance is

$Z=(100-10j)$ ...(II)

On Comparing of equation (I) and (II)

$R = 100$

$X_{L}-X_{C}=-10$

Now, put the value of $X_{C}$

$X_{L=41.66-10$

$X_{L}=31.66\ \Omega$

We need to calculate the inductance

Using formula of inductance

$X_{L}=\omega\times L$

Put the value into the formula

$L=\dfrac{X_{L}}{\omega}$

$L=\dfrac{31.66}{200}$

$L=0.1583\ Henry$

(b). We need to calculate the resonant angular frequency

Using formula of the resonant angular frequency

$angular\ frequency =\dfrac{1}{\sqrt{L\times C}}$

$angular\ frequency =\dfrac{1}{\sqrt{0.1583\times120\times10^{-6}}}$

$angular\ frequency =229.4\ rad/s$

Hence, (a). (i). The reactants are $X_{L} =31.66\ \Omega$ .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.

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3 years ago

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2 years ago

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Georgia [21]

Answer
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Explanation:

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2 years ago

Question 2 (1 point)

Leviafan [203]

The car takes 10.8 seconds to reach a velocity of 42.1 m/s

Explanation:

The motion of the car is at constant acceleration, so we can use the following suvat equation:

$v=u+at$

where

v is the final velocity of the car

u is its initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 15.0 m/s

v = 42.1 m/s

$a=2.50 m/s^2$

Solving for t, we find the time it takes for the car to reach that velocity:

$t=\frac{v-u}{a}=\frac{42.1-15.0}{2.50}=10.8 s$

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