What is considered light? Select ALL that apply

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

6. Is an object moving in uniform circular motion accelerating? Explain.

n200080 [17]

An object moving in a circle is accelerating. Accelerating objects are objects which are changing their velocity.

4 0

3 years ago

Which isotope is least likely to be stable?<br> A) Ca-40 <br> B) Cl-38 <br> C) U-235 <br> D) Xe-136

aksik [14]

Answer:

C) U-235

Explanation:

On Isotope stability, an isotrope will be least stable when Proton and newtron are both odd.

8 0

3 years ago

A sprinter reaches his maximum speed in 2.6 seconds from rest with constant acceleration. He then maintains that speed and finis

Delvig [45]

Answer: maximum speed vmax = 11.42m/s

Explanation:

Given that the sprinter maintained constant acceleration during the first 2.6 seconds.

a = vmax/ta .......1

The distance covered during the acceleration period is;

da = 0.5a(ta)^2 .....2

Substituting equation 1 to 2

da = 0.5(vmax/ta)(ta)^2 = 0.5vmax(ta) .....3

The distance covered during the period of constant speed vmax is;

dv = vmax (tv) ......4

The total distance travelled is

d = da + dv = 100 (Given)

da + dv = 100 ......5

Substituting equation 3 and 4 into 5

0.5vmax(ta) + vmax(tv) = 100

vmax ( 0.5ta +tv) = 100

vmax = 100/(0.5ta + tv) ....6

But,

t = ta + tv

tv = t - ta .......7

Substituting equation 6 into equation 7

vmax = 100/(0.5ta + t - ta)

vmax = 100/(t-0.5ta)

t = 10.06 s

ta = 2.6 s

Substituting the values;

vmax = 100/(10.06 -0.5(2.6))

vmax = 11.42m/s

Note:

ta = acceleration time

tv = constant velocity vmax time

t = overall time

da , dv and d = acceleration, constant velocity and overall distance covered respectively.

8 0

4 years ago

Microphones and loudspeakers are used in an auditorium because the sound waves at the stage compared to the sound waves at the b

MaRussiya [10]

Acoustics (sounds)
Hope the this helps!

6 0

3 years ago