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3 years ago

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In an 8.00 km race, one runner runs at a steady 12.0 km/h and another runs at 14.8 km/h . How far from the finish line is the sl

ower runner when the faster runner finishes the race?

1 answer:

dem82 [27]3 years ago

7 0

2.8km/h see what you do is subtract 12.0 away from 14.8 and that is how you get your answer to the problem.

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12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

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Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust ( $v_{o} = 0\,\frac{m}{s}$ , $a = 30\,\frac{m}{s^{2}}$ , $t_{o} = 0\,s$ , $0\,s\le t< 30\,s$ )

$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

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Answer:

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3 years ago

A ball of mass m is attached to a string of length L. It is being swung in a vertical circle with enough speed so that the strin

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Answer:

$\frac{m}{r}(v_b^2-v_t^2)+2mg$

Explanation:

When the ball is at the bottom position of the vertical circle, the forces acting on the ball are:

- The tension in the string, $T_b$ , upward

- The weight of the ball, $mg$ , downward

The resultant of these forces must be equal to the centripetal force, which points upward as well (towards the center of the circle), so:

$T_b-mg=m\frac{v_b^2}{r}$ (1)

where $v_b$ is the speed of the ball at the bottom of the circle, r the radius of the circle, and m the mass of the ball.

When the ball is at the top position of the vertical circle, the weight still acts downward, however the tension in the string also acts downward, so the equation of the forces becomes:

$T_t+mg=m\frac{v_t^2}{r}$ (2)

where

$T_t$ is the tension in the string in the top position

$v_t$ is the speed of the ball in the top position

By subtracting eq.(2) from eq.(1), we find:

$T_b-mg-(T_t+mg)=m\frac{v_b^2}{r}-m\frac{v_t^2}{r}\\T_b-T_t=\frac{m}{r}(v_b^2-v_t^2)+2mg$

So, this is the difference in tension between the two positions.

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4 years ago

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