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3 years ago

What is the definition of magnitude?

Physics

2 answers:

Margarita [4]3 years ago

8 0

<em>T</em><em>he great size or extent of something. </em>

8_murik_8 [283]3 years ago

4 0

I’m pretty sure it’s size or like how great/big something is

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A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

6 0

3 years ago

Relativistic kinetic energy is conserved in all reference frames. True O False

Free_Kalibri [48]

Answer:

i think it is true

Explanation:

7 0

3 years ago

Now imagine a person dragging a 50 kg box along the ground with a rope, as

ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

= 50 x 9.8

= 490 N

The coefficient of friction,

μ = f/ η

= 30 / 490

= 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0

4 years ago

A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average

Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0

3 years ago

The magnetic field lines outside a bar magnet

n200080 [17]

The lines of magnetic field from a bar magnet form closed lines. By convention, the field direction is taken to be outward from the North pole and in to the South pole of the magnet.

8 0

3 years ago