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2 years ago

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A biker travels at an average speed of 18 km/hr along a 0.30 km straight segment of a bike path. How much time (in hours) does t

he biker take to travel this segment?

1 answer:

lawyer [7]2 years ago

7 0

Answer: 0.016 h

Explanation:

$\text{Average speed} = \frac{\text {Total Distance}}{\text {total time taken}}$

It is given that, biker has an average speed = 18 km/h

Total distance traveled = 0.30 km

Therefore, time taken by biker to travel this distance:

$\Rightarrow \text{total time taken} = \frac{0.30 km}{18 km/h}=0.016 h$

Thus, the biker takes 0.016 hours to travel the segment of 0.30 km at an average speed of 18 km/h.

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a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the fr

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1 year ago

A compact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible fri

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(a) 80 N/m

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The maximum speed of a spring-mass system is given by

$v=\omega A$

where

$\omega$ is the angular frequency

A is the amplitude of the motion

For this system, we have

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

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Substituting into the formula, we find the maximum speed:

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Using the numbers we calculated in part c):

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

$A=0.200 m$

we find immediately the maximum acceleration:

$a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2$

(f) At the position of maximum displacement: $x=\pm 0.200 m$

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$a=\frac{F}{m}$

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

$F=kx$

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

$E=K=\frac{1}{2}mv_{max}^2$

where

m = 2.90 kg is the mass

$v_{max}=1.05 m/s$ is the maximum speed

Solving for E, we find

$E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J$

(h) 0.99 m/s

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$x=\frac{1}{3}(0.200 m)=0.0667 m$

so the elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J$

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$K=E-U=1.60 J-0.18 J=1.42 J$

And from the relationship between kinetic energy and speed, we can find the speed of the system:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s$

(i) 1.84 m/s^2

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$x=\frac{1}{3}(0.200 m)=0.0667 m$

So the restoring force exerted by the spring on the mass is

$F=kx=(80 N/m)(0.0667 m)=5.34 N$

And so, we can calculate the acceleration by using Newton's second law:

$a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2$

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1 year ago