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3 years ago

If your speed changes from 10 km/h to 6 km/h, you have a(n) ______ acceleration.

2 answers:

8 0

If your speed changes from 10 km/h to 6 km/h then
you have an acceleration.

Whether it's a positive or negative one completely depends
on which direction you decided to call the positive direction,
when you started considering your speed and its changes.

If you decided to call the direction in which you're traveling
the positive direction, then a decrease in your speed is a
negative acceleration.

But you could just as easily have said that you're traveling
in the negative direction. If you did that, then a decrease in
your speed would be a positive acceleration.

It's completely up to you, and how you define things.

Leona [35]3 years ago

7 0

If your speed changes from 10 km/h to 6 km/h, you have a(n) a.negative acceleration.

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lorasvet [3.4K]

Answer:

$\omege_A=\omega_M$

$v_M=1.6v_A$

Explanation:

A denotes Alex

M denotes Mary

r = Distance from center

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$\omega_A=\omega_M$

Tangential speed speed is given by

$\dfrac{v_M}{v_A}=\dfrac{r_M\omega_M}{r_A\omega_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{r_M}{r_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{8}{5}\\\Rightarrow \dfrac{v_M}{v_A}=1.6\\\Rightarrow v_M=1.6v_A$

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3 years ago

What is the measurement of six centimeters equal to?

AlladinOne [14]

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Softa [21]

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3 years ago

A stone is thrown from the top of a building 30m high. If it strikes the ground at an angle 45, with what speed was it thrown?

sasho [114]

Answer:

24.2 m/s

Explanation:

The stone strikes the ground at an angle of 45 degrees: this means that its vertical velocity is equal (in magnitude) to its horizontal velocity, in fact:

$tan \theta = \frac{|v_y|}{v_x}\\tan 45^{\circ} = 1 \rightarrow |v_y| = v_x$

The motion along the vertical direction is a uniformly accelerated motion, so we can find the final vertical velocity using the following suvat equation

$v_y^2 -u_y^2 = 2as$

where

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the stone is thrown horizontally)

$a=g=9.8 m/s^2$ is the acceleration of gravity (we take downward as positive direction)

s = 30 m is the vertical displacement

Solving for vy,

$v_y = \sqrt{u_y^2+2as}=\sqrt{0+2(9.8)(30)}=24.2 m/s$

This means that the horizontal velocity is also 24.2 m/s: and since the horizontal velocity is constant during the whole motion (there is no acceleration in the horizontal direction), this means that the stone was thrown exactly at 24.2 m/s.

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3 years ago

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