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adelina 88 [10]

3 years ago

11

A simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top

speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.2 m/s in 2.14 s, what will be his total time?

1 answer:

Trava [24]3 years ago

6 0

Answer:

He will complete the race in total time of T = 10 s

Explanation:

Total distance moved by the sprinter in 2.14 s is given as

$s = \frac{(v_{in} + v_{f})}{2} time$

$s = \frac{(0 + 11.2)}{2} (2.14)$

$s = 11.98 m$

now the distance remaining to move

$d = 100 - 11.98 = 88 m$

now he will move with uniform maximum speed for the remaining distance

so we will have

$time = \frac{d}{v}$

$time = \frac{88}{11.2} = 7.86 s$

so the total time to complete the race is given as

$T = 7.86 + 2.14 = 10 s$

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31.75 m/s

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For first stone:

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Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

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For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

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3 years ago

A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic ener

Mademuasel [1]

The kinetic energy gained by the air molecules is 0.0437 J

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Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

$P =mgh$

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m

P = 0.0441 J

Finally, it will have kinetic energy. It is given by :

$E= \frac{1}{2} mv^{2}$

$E= \frac{1}{2}$ × $1.4$ × 10⁻³ × (0.7)²

E = 0.000343 J

The kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 - 0.0441

= 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

Learn more about kinetic energy here:

brainly.com/question/8101588

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2 years ago

A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel

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Answer:

$4.25 m/s^{2}$

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3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

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$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

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Using the expression of angular acceleration we can find the to then find the torque, that is,

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$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

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With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

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3 years ago