When experimenting, you purposely change the temperature, temperature is the __

A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration

Dimas [21]

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

$x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}$ (1)

Where:

$x_{o}$ - Initial x-position, in meters.

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

$t$ - Time, in seconds.

$a_{x}$ - x-acceleration, in meters per second.

If we know that $x_{o} = 0\,m$ , $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}$

$x(t) = 0.324\,m$

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

$y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}$ (2)

Where:

$y_{o}$ - Initial y-position, in meters.

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

$t$ - Time, in seconds.

$a_{y}$ - y-acceleration, in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o,y} = -3.6\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{y} = 0\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}$

$y(t) = -2.16\,m$

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ( $v_{x}$ ), in meters per second, is calculated by this kinematic formula:

$v_{x}(t) = v_{o,x} + a_{x}\cdot t$ (3)

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-velocity of the skateboarder is:

$v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)$

$v_{x}(t) = 1.08\,\frac{m}{s}$

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

$v_{y} = -3.6\,\frac{m}{s}$

The y-velocity of the skateboard is -3.6 meters per second.

3 0

3 years ago

How long must a pendulum be to have a period of 4.6 5 on Neptune, where g = 11.15 m/s?

Anon25 [30]

Answer:

5.98 m

Explanation:

$T=2\pi \sqrt\frac{L}{g}$

$T^2=4\pi ^2L/g\\L = gT^2/(4\pi^2)$ = 11.15*4.6²/(4π²) = 5.98 m

5 0

2 years ago

Apply what you know about forces to hypothesize how balanced and unbalanced forces acting on a moving car would affect the motio

rosijanka [135]

If an object has a net force acting on it, it will accelerate. The object will speed up, slow down or change direction. An unbalanced force (net force) acting on an objectchanges its speed and/or direction of motion. An unbalanced force is an unopposedforce that causes a change in motion.
thus the car would get its speed, and or direction mixed up

7 0

4 years ago

Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.

olga2289 [7]

Answer:

$NO_{1.499}$