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2 years ago

7

11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of

a car which accelerates from 0 to 40 m/s in 8 seconds?

1 answer:

Marat540 [252]2 years ago

6 0

$a = \frac{v - u}{t}$

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

$\frac{10 - 0}{3} = 3.333333....$

approximately 3.33 m/s^2

the acceleration of the car is

$\frac{40 - 0 }{8} = 5.0$

5.0 m/s^2

$5.0 > 3.33 \\ so \: the \: answer \: is \: no$

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Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove

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Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

$ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2$

$kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2$

Since the masses are equals:

$m=ma=mb$

For the known velocities, the kinetics energies result:

$ki=0.5*mVa_{i}^2$

$ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m$

$kf=0.5*mbVb_{f}^2$

$kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m$

The lost energy in the collision is the difference between the initial and final kinectic energies:

$kl=ki-kf$

$kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m$

Finally the relation between the lost and the initial kinetic energy:

$kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m$

$kl/ki = 24/49=0.49$

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3 years ago

A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha

ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, $L=400\ \mu H=400\times 10^{-6}\ H$

Capacitance of parallel LCR circuit, $C=0.3\ \mu F=0.3\times 10^{-6}\ F$

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

$f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }$

$f=14528.79\ Hz$

or

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

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garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

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Explanation:

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